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19 tháng 10 2023

\(\dfrac{-8}{3}+\dfrac{2}{11}-\dfrac{8}{3}:\dfrac{11}{9}\)

\(=\dfrac{-8}{3}+\dfrac{2}{11}+\dfrac{-8}{3}\cdot\dfrac{9}{11}\)

\(=\dfrac{-8}{3}+\dfrac{2}{11}-\dfrac{24}{11}\)

\(=\dfrac{-8}{3}-\dfrac{22}{11}\)

\(=\dfrac{-8}{3}-2\)

\(=\dfrac{-8}{3}-\dfrac{6}{3}\)

\(=-\dfrac{14}{3}\)

19 tháng 10 2023

-8/3 +2/11 -8/3 :11/9
= -88/33 +6/33 -72/33
=-154/33=-14/3

AH
Akai Haruma
Giáo viên
18 tháng 9 2018

Lời giải:

a)

\(\frac{\frac{2}{3}-\frac{2}{5}+\frac{2}{7}-\frac{2}{9}+\frac{2}{11}}{\frac{8}{3}-\frac{8}{5}+\frac{8}{7}-\frac{8}{9}+\frac{8}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}{8\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}\) \(=\frac{2}{8}=\frac{1}{4}\)

b)

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{50}-1\right)\left(\frac{1}{51}-1\right)\)

\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}....\frac{1-50}{50}.\frac{1-51}{2}=\frac{(-1)(-2)(-3)...(-49)(-50)}{2.3.4....50.51}\)

\(=\frac{(-1)^{50}.1.2.3....49.50}{2.3.4...50.51}=\frac{1}{51}\)

18 tháng 9 2023

a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)

\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)

b) 

\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)

\(=-14\)

c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)

\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)

12 tháng 11 2023

\(\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right):\dfrac{5}{9}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right):\dfrac{5}{9}\right]\cdot\dfrac{11}{235}\)

\(=\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right)\cdot\dfrac{9}{5}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{9}{5}\right]\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot\left(\dfrac{-3}{8}+\dfrac{11}{23}+\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot\left[\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{11}{23}+\dfrac{12}{23}\right)\right]\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot\left(-1+1\right)\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot0\cdot\dfrac{11}{235}\)

\(=0\)

8 tháng 4 2023

\(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{8}+0,7}\\ =\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\\ =\dfrac{2}{7}-\dfrac{2\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}\\ =\dfrac{2}{7}-\dfrac{2}{7}=0\)

8 tháng 4 2023

phân số cuối là \(\dfrac{2}{7}-\dfrac{2\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}\) nha :vv

20 tháng 12 2022

a: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{3}{2}=\dfrac{9-10+18}{12}=\dfrac{17}{12}\)

b: \(=\left(\dfrac{1}{9}+\dfrac{6}{9}\right)^2-\dfrac{1}{3}=\dfrac{49}{81}-\dfrac{27}{81}=\dfrac{22}{81}\)

c; \(=\dfrac{5}{11}\left(-\dfrac{3}{7}-\dfrac{5}{7}\right)+\dfrac{-8}{7}\cdot\dfrac{6}{11}=\dfrac{-8}{7}\left(\dfrac{5}{11}+\dfrac{6}{11}\right)=-\dfrac{8}{7}\)

d: \(=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{1}{2}\)

22 tháng 10 2021

Bài 1:

a) \(\dfrac{19}{12}+\left|\dfrac{-5}{2}\right|+\left(\dfrac{3}{2}\right)^2=\dfrac{19}{12}+\dfrac{5}{2}+\dfrac{9}{4}\)

\(=\dfrac{19+5.6+9.3}{12}=\dfrac{76}{12}=\dfrac{19}{3}\)

b) \(\dfrac{2}{11}.\dfrac{16}{9}-\dfrac{2}{11}.\dfrac{7}{9}=\dfrac{2}{11}\left(\dfrac{16}{9}-\dfrac{7}{9}\right)=\dfrac{2}{11}.1=\dfrac{2}{11}\)

Bài 2:

Áp dụng t/c dtsbn:

\(\dfrac{a}{8}=\dfrac{b}{3}=\dfrac{a-b}{8-3}=\dfrac{55}{5}=11\)

\(\Rightarrow\left\{{}\begin{matrix}x=11.8=88\\b=11.3=33\end{matrix}\right.\)

26 tháng 7 2018

a. = \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}-\dfrac{-3}{8}\right\}\)

= \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}+\dfrac{3}{8}\right\}\)

= \(\dfrac{-1}{24}-\dfrac{5}{8}\)

= \(\dfrac{-2}{3}\)

b. = \(12\dfrac{7}{88}-3\dfrac{5}{11}\)

= \(8\dfrac{5}{8}\)

c. = \(\dfrac{-28}{9}+\dfrac{-413}{9}\)

= \(-49\)

d. = \(\dfrac{8}{35}:\dfrac{2}{11}+\dfrac{-8}{35}:\dfrac{2}{11}\)

= \(\dfrac{2}{11}:\left(\dfrac{8}{35}+\dfrac{-8}{35}\right)\)

= 0

\(A=\dfrac{\dfrac{3}{11}+\dfrac{3}{3}-\dfrac{3}{7}}{\dfrac{9}{11}+\dfrac{9}{3}-\dfrac{9}{7}}-\dfrac{\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{8}}{\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}+\dfrac{7}{16}}\)

\(=\dfrac{1}{3}-1:\dfrac{7}{2}=\dfrac{1}{3}-\dfrac{2}{7}=\dfrac{1}{21}\)

24 tháng 9 2021

\(\dfrac{x+4}{8}+\dfrac{x+3}{9}=\dfrac{x+2}{10}+\dfrac{x+1}{11}\)

\(\Leftrightarrow\left(\dfrac{x+4}{8}+1\right)+\left(\dfrac{x+3}{9}+1\right)=\left(\dfrac{x+2}{10}+1\right)+\left(\dfrac{x+1}{11}+1\right)\)

\(\Leftrightarrow\dfrac{x+12}{8}+\dfrac{x+12}{9}-\dfrac{x+12}{10}-\dfrac{x+12}{11}=0\)

\(\Leftrightarrow\left(x+12\right)\left(\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}\right)=0\)

\(\Leftrightarrow x=-12\)( do \(\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}\ne0\))

24 tháng 9 2021

\(\dfrac{x+4}{8}+\dfrac{x+3}{9}=\dfrac{x+2}{10}+\dfrac{x+1}{11}\)

\(\dfrac{x+4}{8}+1+\dfrac{x+3}{9}+1=\dfrac{x+2}{10}+1+\dfrac{x+1}{11}+1\)

\(\dfrac{x+12}{8}+\dfrac{x+12}{9}=\dfrac{x+12}{10}+\dfrac{x+12}{11}\)

\(\dfrac{x+12}{8}+\dfrac{x+12}{9}-\dfrac{x+12}{10}-\dfrac{x+12}{11}=0\)

\(\Rightarrow\left(x+12\right).\left(\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}\right)=0\)

Vì \(\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}\ne0\) nên \(x+12=0\)

\(\Rightarrow x=-12\)