1.Tính giá trị biểu thức sau :a, \(\left(\frac{3}{4}+\frac{3}{2}\right):\frac{7}{4}-\frac{3}{4}\) b,... - Olm

K

Khách

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VD

vu duc huy

27 tháng 2 2021

1.Tính giá trị biểu thức sau :

a, \(\left(\frac{3}{4}+\frac{3}{2}\right):\frac{7}{4}-\frac{3}{4}\) b, \(\left(-5\right)^2\cdot\frac{7}{45}+\left(-5\right)^2\cdot\frac{11}{45}\)

2

2

༺༒༻²ᵏ⁸

27 tháng 2 2021

\(a,\left(\frac{3}{4}+\frac{3}{2}\right):\frac{7}{4}-\frac{3}{4}\)

\(=\left(\frac{3}{4}+\frac{6}{4}\right).\frac{4}{7}-\frac{3}{4}\)

\(=\frac{9}{4}.\frac{4}{7}-\frac{3}{4}\)

\(=\frac{9}{7}-\frac{3}{4}\)

\(=\frac{36}{28}-\frac{21}{28}\)

\(=\frac{15}{28}\)

\(b,\left(-5\right)^2.\frac{7}{45}-\left(-5\right)^2.\frac{11}{45}\)

\(=\left(-5\right)^2.\left(\frac{7}{45}-\frac{11}{45}\right)\)

\(=\left(-5\right)^2.\frac{-4}{45}\)

\(=25.\frac{-4}{45}\)

\(=\frac{-100}{45}\)

\(=\frac{20}{9}\)

#Chúc em học tốt

NT

Nguyễn Thùy Chi

27 tháng 2 2021

BẰNG 1+56[90.;KO90

Dưới đây là một vài câu hỏi có thể liên quan tới câu hỏi mà bạn gửi lên. Có thể trong đó có câu trả lời mà bạn cần!

NT

Nhân Tư

2 tháng 7 2016

\(\frac{AB^4}{AC^4}=\frac{\left(AB^2\right)^2}{\left(AC^2\right)^2}=\frac{\left(BH\cdot BC\right)^2}{\left(CH\cdot BC\right)^2}=\frac{BC^2}{CH^2}=\frac{BE\cdot AB}{CF\cdot AC}=\frac{BE}{CF}\cdot\frac{AB}{AC}\Leftrightarrow\frac{AB^3}{AC^3}=\frac{BE}{CF}\)

3

B

Bloom

2 tháng 7 2016

đây đau phải la lớp1

H

hungbck5

2 tháng 7 2016

khong phai toan lop 1

V

vinh

19 tháng 8 2019

e,\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{42}\right)\)\(\Rightarrow A=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}=4-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)\(\Rightarrow...

2

BV

Bò Vinamilk 3 không (Hội Con 🐄)

19 tháng 8 2019

Đúng rùi bn

BV

Bò Vinamilk 3 không (Hội Con 🐄)

19 tháng 8 2019

BN mún hỏi j vậy, đây k phải câu hỏi, mà có thì phải là toán lớp 6

K

KAl(SO4)2·12H2O

5 tháng 10 2019

1) Tính:a) \(\frac{3}{5}+\left(-\frac{1}{4}\right)\)b) \(\left(-\frac{5}{18}\right)\left(-\frac{9}{10}\right)\)c) \(4\frac{3}{5}:\frac{2}{5}\)2) Tìm x:a)\(\frac{12}{x}=\frac{3}{4}\)b) \(x:\left(\frac{-1}{3}\right)^3=\left(\frac{-1}{3}\right)^2\)c) \(\frac{-11}{12}.x+0,25=\frac{5}{6}\)d) \(\left(x-1\right)^5=-32\)3) Cho |m| = -3, tìm m:4) Các cạnh của một tam giác có số đo tỉ lệ với các số 3; 4; 5. Tính cạnh của tam giác biết chu vi của nó là 13,2...

3

BA

Bùi Anh Tuấn

5 tháng 10 2019

Bài 1

\(a,\frac{3}{5}+\left(-\frac{1}{4}\right)=\frac{7}{20}\)

\(b,\left(-\frac{5}{18}\right)\cdot\left(-\frac{9}{10}\right)=\frac{1}{4}\)

\(c,4\frac{3}{5}:\frac{2}{5}=\frac{23}{5}\cdot\frac{5}{2}=\frac{23}{2}\)

Bài 2

\(a,\frac{12}{x}=\frac{3}{4}\Rightarrow3x=12\cdot4\)

\(\Rightarrow3x=48\)

\(\Rightarrow x=16\)

\(b,x:\left(-\frac{1}{3}\right)^3=\left(-\frac{1}{3}\right)^2\)

\(\Rightarrow x=\left(-\frac{1}{3}\right)^2\cdot\left(-\frac{1}{3}\right)^3=\left(-\frac{1}{3}\right)^5\)

\(\Rightarrow x=-\frac{1}{243}\)

\(c,-\frac{11}{12}\cdot x+0,25=\frac{5}{6}\)

\(\Rightarrow-\frac{11}{12}x=\frac{5}{6}-\frac{1}{4}=\frac{7}{12}\)

\(\Rightarrow x=\frac{7}{12}:\left(-\frac{11}{12}\right)\)

\(\Rightarrow x=-\frac{7}{11}\)

\(d,\left(x-1\right)^5=-32\)

\(\left(x-1\right)^5=-2^5\)

\(x-1=-2\)

\(x=-2+1=-1\)

Bài 3

\(\left|m\right|=-3\Rightarrow m\in\varnothing\)

Bài 3

Gọi 3 cạnh của tam giác lần lượt là a;b;c ( a,b,c>0)

Ta có

\(a+b+c=13,2\)

\(\frac{a}{3};\frac{b}{4};\frac{c}{5}\)

Ap dụng tính chất DTSBN ta có

\(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{a+b+c}{3+4+5}=\frac{13,2}{12}=\frac{11}{10}\)

\(\hept{\begin{cases}\frac{a}{3}=\frac{11}{10}\\\frac{b}{4}=\frac{11}{10}\\\frac{c}{5}=\frac{11}{10}\end{cases}}\Rightarrow\hept{\begin{cases}a=\frac{33}{10}\\b=\frac{44}{10}=\frac{22}{5}\\c=\frac{55}{10}=\frac{11}{2}\end{cases}}\)

Vậy 3 cạnh của tam giác lần lượt là \(\frac{33}{10};\frac{22}{5};\frac{11}{2}\)

TL

๖ۣۜƝƘ☆❖TྂRí❖TIêNྂᴾᴿᴼAK❖47シRồng lửa

5 tháng 10 2019

a)\(\frac{3}{5}+\left(-\frac{1}{4}\right)\)

\(=\frac{3}{5}-\frac{1}{4}\)

\(=\frac{12}{20}-\frac{5}{20}=\frac{7}{20}\)

b)\(\left(-\frac{5}{18}\right)\left(-\frac{9}{10}\right)\)

\(=\frac{\left(-5\right)\left(-9\right)}{18.10}\)

\(=\frac{\left(-1\right)\left(-1\right)}{2.2}=\frac{1}{4}\)

c)\(4\frac{3}{5}:\frac{2}{5}\)

\(=\frac{23}{5}:\frac{2}{5}\)

\(=\frac{23}{5}.\frac{5}{2}\)

\(=\frac{23.1}{1.2}=\frac{23}{2}\)

1/

a)\(\frac{12}{x}=\frac{3}{4}\)

\(\Rightarrow x.3=12.4\)

\(\Rightarrow x.3=48\)

\(\Rightarrow x=48:3=16\)

b)\(x:\left(\frac{-1}{3}\right)^3=\left(\frac{-1}{3}\right)^2\)

\(x=\left(\frac{-1}{3}\right)^2.\left(\frac{-1}{3}\right)^3\)

\(x=\frac{\left(-1\right)^2}{3^2}.\frac{\left(-1\right)^3}{3^3}\)

\(x=\frac{1}{9}.\frac{-1}{27}=-\frac{1}{243}\)

M

manh

15 tháng 3 2019

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}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}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}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}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{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$v

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Đ𝐚𝐧𝐧 𝐋ê

15 tháng 3 2019

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ミ★ 🆂🆄🅽 ★彡

15 tháng 3 2019

I. Nội qui tham gia "Giúp tôi giải toán"

1. Không đưa câu hỏi linh tinh lên diễn đàn, chỉ đưa các bài mà mình không giải được hoặc các câu hỏi hay lên diễn đàn;

2. Không trả lời linh tinh, không phù hợp với nội dung câu hỏi trên diễn đàn.

3. Không "Đúng" vào các câu trả lời linh tinh nhằm gian lận điểm hỏi đáp.

Các bạn vi phạm 3 điều trên sẽ bị giáo viên của Online Math trừ hết điểm hỏi đáp, có thể bị khóa tài khoản hoặc bị cấm vĩnh viễn không đăng nhập vào trang web.

PT

pham thi thu trang

18 tháng 4 2018

1

PG

Phan Gia Huy

9 tháng 1 2021

bạn trung học hay tiểu học vậy

LD

l҉o҉n҉g҉ d҉z҉

25 tháng 7 2020

Câu hỏi hay

1.Chứng minh rằng :

\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+b+c+d\)với \(a\ge-1;b\ge-4;c\ge2;d>3\)

2. Chứng minh rằng :

\(\frac{a^2}{b^5}+\frac{b^2}{c^5}+\frac{c^2}{d^5}+\frac{d^2}{a^5}\ge\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)với \(a,b,c,d>0\)

59

HF

HD Film

25 tháng 7 2020

Câu 1:
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+1+b+4+c-2+d-3=a+b+c+d\)

Dấu = xảy ra khi a = -1; b = -4; c = 2; d= 3

PM

Phùng Minh Quân

25 tháng 7 2020

\(\frac{a^2}{b^5}+\frac{1}{a^2b}\ge\frac{2}{b^3}\)\(\Leftrightarrow\)\(\frac{a^2}{b^5}\ge\frac{2}{b^3}-\frac{1}{a^2b}\)

\(\frac{2}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\)\(\Leftrightarrow\)\(\frac{1}{a^2b}\le\frac{2}{3a^3}+\frac{1}{3b^3}\)

\(\Rightarrow\)\(\Sigma\frac{a^2}{b^5}\ge\Sigma\left(\frac{5}{3b^3}-\frac{2}{3a^3}\right)=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)

LT

lê thị ngọc thảo

27 tháng 3 2017

a) \(x^3 + 1 = (x + 1)(x^2 - x + 1)\) \(x^9 + x^7 - 3x^2 - 3 = x^7(x^2 + 1) - 3(x^2 + 1) = (x^2 + 1)(x^7 - 3)\).Điều kiện của x để giá trị của biểu thức Q xác định là \(x \neq -1, x^7 \neq 3, x \neq -3, x \neq 4\).b) \(Q = \left[\frac{x^7 -3}{x^3 + 1}.\frac{(x - 1)(x + 1)(x^2 - x + 1)}{(x^7 - 3)(x^2 + 1)} + 1 - \frac{2(x + 6)}{x^2 + 1}\right].\frac{(2x + 1)^2}{(x + 3)(4 - x)}\) \(= \left[\frac{x^7 - 3}{x^3 + 1}.\frac{(x - 1)(x^3 + 1)}{(x^7 - 3)(x^2 + 1)} + 1 -...

2

NH

Nguyễn hiểu minh

27 tháng 3 2017

2 nha bn

PM

Phạm Minh Tuệ

5 tháng 8 2021

lớp 1 căng đét

T

Tuấn

30 tháng 7 2016

a,\(8x^3-12x^2+6x-5=0\Leftrightarrow8\left(x^3-\frac{3}{2}x^2+\frac{3}{4}x-\frac{1}{8}\right)-4=0\)
\(\Leftrightarrow8\left(x-\frac{1}{2}\right)^3=4\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\frac{1}{2}\Leftrightarrow x=\frac{1}{\sqrt[3]{2}}+\frac{1}{2}\)

2

T

Tuấn

30 tháng 7 2016

b, 3x^3+3x^2+3x+1=0<=>2x^3+(x+1)^3=0<=> .
Hằng đẳng thức đi bác

KN

ka nekk

26 tháng 2 2022

đây đích thực có phải lớp 1 ko bn?

LH

Lưu Hương Lý

3 tháng 12 2017

\(\left(\frac{5}{4}-\frac{2}{5}\right).\frac{2012}{2013}+\left(\frac{3}{4}-\frac{3}{5}\right).\frac{2012}{2013}\)

2

M

Mafia

3 tháng 12 2017

\(=\left(\frac{5}{4}-\frac{2}{5}+\frac{3}{4}-\frac{3}{5}\right).\frac{2012}{2013}\)

\(=\left(\frac{8}{4}-\frac{5}{5}\right).\frac{2012}{2013}\)

\(=\left(2-1\right).\frac{2012}{2013}\)

\(=\frac{2012}{2013}\)

QA

Quỳnh Anh Lưu

3 tháng 12 2017

\(=\frac{2012}{2013}.\left(\frac{5}{4}-\frac{2}{5}+\frac{3}{4}-\frac{3}{5}\right)=\frac{2012}{2013}.\left(\frac{5}{4}+\frac{3}{4}-\frac{2}{5}-\frac{3}{5}\right)=\frac{2012}{2013}.\left(\frac{8}{4}-\frac{5}{5}\right)=\frac{2012}{2013}.\left(1-1\right)=\frac{2012}{2013}.0\)