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Nên đợi ai đó giải hết 2 3 bài xong rồi mới đăng tiếp những bài còn lại, chứ dài vậy giải hơi nản =)))
Bài 1:
1, \(13\frac{2}{5}-\left(\frac{18}{32}-2\frac{6}{10}\right)\)
\(=\frac{67}{5}-\left(\frac{9}{16}-\frac{13}{5}\right)\)(Chuyển hỗn số thành p/số và rút gọn hai số trong ngoặc luôn)
\(=\frac{67}{5}-\left(\frac{-163}{80}\right)\)
\(=\frac{246}{16}\)
2, \(22.4\frac{5}{7}-\left(8.91+1,09\right)\)(Phần 2 viết vầy có đúng không vậy ? Nếu sai thì kêu chị sửa nhé)
\(=22.\frac{33}{7}-10\)
\(=\frac{726}{7}-10\)
\(=\frac{656}{7}\)
3, Chỗ ''3 phần 10 phần 2'' là sao :v ?
4, \(5\frac{2}{7}.\frac{8}{11}+5\frac{2}{7}.\frac{5}{11}-5\frac{2}{7}.\frac{2}{11}\)
\(=\frac{37}{7}.\frac{8}{11}+\frac{37}{7}.\frac{5}{11}-\frac{37}{7}.\frac{2}{11}\)(Chuyển hỗn số thành p/số)
\(=\frac{37}{7}.\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)(Dùng tính chất phân phối)
\(=\frac{37}{7}.\frac{11}{11}\)
\(=\frac{37}{7}.1=\frac{37}{7}\)
\(a,\frac{3}{17}+\frac{-5}{13}+\frac{-18}{35}+\frac{14}{17}+\frac{17}{-35}\)
=\(-\frac{5}{13}+\left(\frac{3}{17}+\frac{14}{17}\right)+\left(\frac{-18}{35}+\frac{-17}{35}\right)\)
= \(-\frac{5}{13}+1+\left(-1\right)\)
=\(-\frac{5}{13}\)
\(b,\frac{-3}{8}.\frac{1}{6}+\frac{3}{-8}.\frac{5}{6}+\frac{-10}{6}\)
=\(\frac{-3}{8}.\left(\frac{1}{6}+\frac{5}{6}\right)+\frac{-10}{6}\)
=\(\frac{-3}{8}.1+\frac{-10}{6}\)
=\(-\frac{49}{24}\)
\(c,\frac{-4}{11}.\frac{5}{15}.\frac{11}{-4}\)
=\(\left(\frac{-4}{11}.\frac{11}{-4}\right).\frac{1}{3}\)
=\(1.\frac{1}{3}=\frac{1}{3}\)
\(d,\frac{13}{8}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-25\%.\frac{1}{2}\)
=\(\frac{13}{8}+\frac{1}{8}:\left(\frac{3}{4}-\frac{1}{2}\right)-\frac{1}{4}.\frac{1}{2}\)
=\(\frac{13}{8}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}\)
=\(\frac{13}{8}+\frac{1}{2}+\frac{-1}{8}\)
=\(\left(\frac{13}{8}+\frac{-1}{8}\right)+\frac{1}{2}\)
=\(\frac{3}{2}+\frac{1}{2}=2\)
\(e,\frac{-1}{2^2}-\left(-2\right)^2-5\)
=\(\frac{-1}{4}-4-5\)
=\(-\frac{37}{4}\)
\(f,\frac{121}{3}-\frac{5}{7}:\left(24-\frac{23}{57}\right)\)
=\(\frac{121}{3}-\frac{5}{7}:\frac{1345}{57}\)
=\(\frac{121}{3}-\frac{57}{1883}\)
\(\approx40,4\)
1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅
3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1
5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)
6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅
7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅
8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1
9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)
\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
Câu 3, 4 tương tự nhé.
\(\frac{3}{4}+\frac{-1}{3}+\frac{-5}{18}\)
\(=\frac{5}{12}+\frac{-5}{18}\)
\(=\frac{5}{36}\)
1: Rút gọn
a) \(-\dfrac{33}{51}=\dfrac{-33:3}{51:3}=\dfrac{-11}{17}\)
b) \(\dfrac{156}{-168}=\dfrac{-156}{168}=\dfrac{-156:12}{168:12}=\dfrac{-13}{14}\)
c) \(\dfrac{-75}{-100}=\dfrac{75}{100}=\dfrac{75:25}{100:25}=\dfrac{3}{4}\)
2: Quy đồng mẫu:
a) \(-\dfrac{3}{8}\) và \(\dfrac{5}{4}\)
MSC: 8
\(-\dfrac{3}{8}=\dfrac{-3}{8}\)
\(\dfrac{5}{4}=\dfrac{5\cdot2}{4\cdot2}=\dfrac{10}{8}\)
b) \(\dfrac{-7}{6}\) ; \(\dfrac{5}{12}\) và \(-\dfrac{5}{6}\)
MSC: 12
\(-\dfrac{7}{6}=\dfrac{-7\cdot2}{6\cdot2}=\dfrac{-14}{12}\)
\(\dfrac{5}{12}=\dfrac{5}{12}\)
\(\dfrac{-5}{6}=\dfrac{-5\cdot2}{6\cdot2}=\dfrac{-10}{12}\)
Ta có : \(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)
\(\frac{1}{5^2}=\frac{1}{5.5}< \frac{1}{4.5}\)
\(\frac{1}{6^2}=\frac{1}{6.6}< \frac{1}{5.6}\)
...
\(\frac{1}{100^2}=\frac{1}{100.100}< \frac{1}{99.100}\)
\(\Rightarrow\)K<\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
K<\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
K<\(\frac{1}{3}-\frac{1}{100}< \frac{1}{3}\)
\(\Rightarrow K< \frac{1}{3}\) (1)
Ta có : \(\frac{1}{4^2}=\frac{1}{4.4}=\frac{1}{16}\)
\(\frac{1}{5^2}=\frac{1}{5.5}>\frac{1}{5.6}\)
\(\frac{1}{6^2}=\frac{1}{6.6}>\frac{1}{6.7}\)
...
\(\frac{1}{99^2}=\frac{1}{99.99}>\frac{1}{99.100}\)
\(\frac{1}{100^2}=\frac{1}{100.100}>\frac{1}{100.101}\)
\(\Rightarrow K>\frac{1}{16}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}+\frac{1}{100.101}\)
K>\(\frac{1}{16}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)
K>\(\frac{1}{16}+\frac{1}{5}-\frac{1}{101}>\frac{1}{5}\) (2)
Từ (1) và (2)
\(\Rightarrow\frac{1}{5}< K< \frac{1}{3}\)
Vậy \(\frac{1}{5}< K< \frac{1}{3}.\)
e , l ko làm
\(a,\frac{3}{7}+\left(-\frac{5}{2}\right)+\left(-\frac{3}{5}\right)\)
= \(\frac{30-175-42}{70}\)
= \(-\frac{187}{70}\)
\(b,3,5-\left(-\frac{2}{7}\right)\)
= \(\frac{7}{2}+\frac{2}{7}\)
= \(\frac{53}{14}\)
\(c,\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\)
= \(\frac{56+20-49}{70}\)
= \(\frac{27}{70}\)
\(d,\frac{2}{3}.\frac{5}{2}-\frac{3}{4}.\frac{2}{3}\)
\(=\frac{2}{3}.\left(\frac{5}{2}-\frac{3}{4}\right)\)
\(=\frac{2}{3}.\frac{7}{4}\)
\(=\frac{7}{6}\)
\(f,-\frac{6}{21}.\frac{3}{2}\)
= \(-\frac{3}{7}\)
\(g,-3.\left(-\frac{7}{12}\right)\)
\(=\frac{7}{4}\)
\(h,\left(\frac{11}{12}:\frac{33}{16}\right).\frac{3}{5}\)
\(=\frac{4}{9}.\frac{3}{5}\)
\(=\frac{4}{15}\)
\(i,12.\left(-\frac{2}{3^2}\right)+\frac{4}{3}\)
= \(-\frac{24}{9}+\frac{4}{3}\)
\(=-\frac{4}{3}\)
\(j,\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4^2}\right)\)
= \(\frac{4}{5}.\left(\frac{56+1}{16}\right)\)
\(=\frac{4}{5}.\frac{57}{16}\)
= \(\frac{57}{20}\)
\(k,\frac{5^4}{25^5}.\frac{20^4}{4^5}\)
\(=\frac{5^4}{\left(5^2\right)^5}.\frac{\left(4.5\right)^4}{4^5}\)
= \(\frac{5^4}{5^{10}}.\frac{4^4.5^4}{4^5}\)
= \(\frac{1}{5^6}.\frac{5^4}{4}\)
=\(\frac{5^4}{5^6.4}\)
=\(\frac{1}{5^2.4}\)
=\(\frac{1}{100}\)
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