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6 tháng 8 2021

a) \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)

b) \(B=2\dfrac{3}{11}.1\dfrac{1}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{13}{12}.\dfrac{-11}{5}=-\dfrac{65}{12}\)

c) \(C=\left(\dfrac{3}{4}-0,2\right)\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\left(\dfrac{-2}{5}\right)=\dfrac{-11}{50}\)

6 tháng 8 2021

A = 2/3 + -1/3

    = 1/3

B = 25/11 . 13/12 . (-2,2)

    = 325/132 . (-2,2)

    = -65/12

C = 11/20 . -2/5

    = -11/50

Chúc bạn học tốt!! ^^

    = -

9 tháng 6 2017

Sau khi thực hiện phép tính ta được kết quả các giá trị:

\(A=\dfrac{1}{3}\) \(B=-5\dfrac{5}{12}\) \(C=-0,22\)

Sắp xếp: \(-5\dfrac{5}{12}< -0,22< \dfrac{1}{3}\) tức là \(B< C< A\)

28 tháng 8 2017

Khi tính xong giá trị biểu thức A , B và C ta được kết quả như sau :

\(A=\dfrac{1}{3}\) ; \(B=-5\dfrac{5}{12}\); \(C=-0,22\)

Sắp xếp : \(B< C< A\)\(\left(-5\dfrac{5}{12}< -0,22< \dfrac{1}{3}\right)\)

21 tháng 6 2023

\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)

\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)

\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)

\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)

\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)

\(=7-\dfrac{26}{5}\)

\(=\dfrac{9}{5}\)

\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)

\(=\dfrac{2}{3}+\dfrac{21}{8}\)

\(=\dfrac{79}{24}\)

\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)

\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)

\(=\dfrac{31}{4}:\dfrac{49}{8}\)

\(=\dfrac{62}{49}\)

\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)

 

a: \(A=\dfrac{3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}}{3^{12}\cdot5^6+5^6\cdot3^{12}}\)

\(=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{2\cdot3^{12}\cdot5^6}\)

\(=\dfrac{3^{13}\cdot5^4\cdot\left(3-1\right)}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)

c: \(C=\dfrac{\dfrac{27}{64}+\dfrac{125}{64}-5\cdot\dfrac{16-15}{12}}{\dfrac{25}{64}+\dfrac{4}{9}-\dfrac{5}{6}}\)

\(=\dfrac{47}{24}:\dfrac{1}{576}=47\cdot24=1128\)

 

AH
Akai Haruma
Giáo viên
10 tháng 10 2023

Lời giải:
a. $=0,16-(-0,064).(-3)=0,16-0,192=-0,032$
b. $=(1\frac{3}{4})^2(1\frac{3}{4}-1)+1=(1\frac{3}{4})^2.\frac{3}{4}+1$

$=\frac{147}{64}+1=\frac{211}{64}$

c.

$=(\frac{2}{3})^3-4(\frac{-7}{4})^2-(\frac{2}{3})^3$

$=-4(\frac{-7}{4})^2=\frac{-49}{4}$

11 tháng 10 2023

a) = 0,16 - 0,064 . (-3)

= 0,16 + 0,192

= 0,352

b) = (7/4)³ - (7/4)² + 1

= 343/64 - 49/16 + 1

= 147/64 + 1

= 211/64

c) = 8/27 - 4.(-7/4)² - 8/27

= -4.49/16

= -49/4

a: \(=\dfrac{2}{3}+\dfrac{38}{7}\cdot\dfrac{1}{2}-\dfrac{9}{7}+\dfrac{1}{12}\cdot16\)

\(=\dfrac{2}{3}+\dfrac{4}{3}+\dfrac{19}{7}-\dfrac{9}{7}=2+\dfrac{10}{7}=\dfrac{24}{7}\)

b: \(=\dfrac{11}{4}\cdot\dfrac{-2}{5}-\dfrac{11}{4}\cdot\dfrac{8}{5}+\dfrac{11}{4}\cdot\dfrac{-6}{5}\)

\(=\dfrac{11}{4}\cdot\dfrac{-16}{5}=\dfrac{-44}{5}\)

6 tháng 12 2017

B = .................

Xét thừa số 63.1,2 - 21.3,6 = 0 nên B = 0

\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)

\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{2\left(\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}\right)}\right|\)

\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{1}{2}\right|=\dfrac{1}{18}+\dfrac{9}{10}=\dfrac{43}{45}\)

6 tháng 12 2017

Mình làm câu 1,2 trước, câu 3 sau

Câu 1:

\(\sqrt{x^2}=0\)

=> \(\left(\sqrt{x^2}\right)^2=0^2\)

\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)

Câu 2:

\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)

\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)

\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)

\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)

21 tháng 9 2023

\(a,0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{5}{6}+\dfrac{39}{35}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\left(\dfrac{5}{6}-\dfrac{1}{6}\right)+\left(\dfrac{39}{35}-\dfrac{4}{35}\right)\\ =\dfrac{2}{3}+1\\ =\dfrac{4}{3}.\)

\(b,\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(-6-\dfrac{7}{4}+\dfrac{3}{2}\right)\\ =3-\dfrac{1}{4}+\dfrac{2}{3}-5-\dfrac{1}{3}+\dfrac{6}{5}+6+\dfrac{7}{4}-\dfrac{3}{2}\\ =\left(3-5+6\right)+\left(-\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{6}{5}+\dfrac{7}{4}\right)\\ =4-\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{5}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{17}{6}+\dfrac{59}{20}\\ =\dfrac{347}{60}.\)

\(c,\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\left(\dfrac{1}{3}-\dfrac{2}{9}\right)+\left(\dfrac{3}{4}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\dfrac{1}{64}\\ =\dfrac{1}{9}+\dfrac{13}{18}+\dfrac{2}{3}+\dfrac{1}{64}\\ =\dfrac{3}{2}+\dfrac{1}{64}\\ =\dfrac{65}{64}.\)

21 tháng 9 2023

câu cuối có sai ko bn?

20 tháng 10 2023

a: \(\left(\dfrac{5}{9}-\dfrac{\sqrt{9}}{12}\right):\dfrac{3}{4}+\dfrac{11}{3}:\dfrac{3}{4}\)

\(=\left(\dfrac{5}{9}-\dfrac{3}{12}\right)\cdot\dfrac{4}{3}+\dfrac{11}{3}\cdot\dfrac{4}{3}\)

\(=\left(\dfrac{5}{9}-\dfrac{1}{4}+\dfrac{11}{3}\right)\cdot\dfrac{4}{3}\)

\(=\dfrac{20-9+132}{36}\cdot\dfrac{4}{3}\)

\(=\dfrac{143}{3}\cdot\dfrac{1}{9}=\dfrac{143}{27}\)

b: \(\left(0.\left(3\right)+\dfrac{\left|-2\right|}{3}\right):\dfrac{\sqrt{25}}{4}-\left(2^3+3^2\right)^0\)

\(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\cdot\dfrac{4}{5}-1\)

\(=\dfrac{4}{5}-1=-\dfrac{1}{5}\)