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25 tháng 3 2022

x = 3/4

25 tháng 3 2022

\(x=\dfrac{5}{8}+\dfrac{1}{8}\)

\(x=\dfrac{6}{8}=\dfrac{3}{4}\)

25 tháng 3 2022

\(x=\dfrac{3}{5}-\dfrac{7}{8}=-\dfrac{11}{40}\)

25 tháng 3 2022

x = 3/5 - 7/8

x =  -11/40

25 tháng 3 2022

\(=>\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)=\dfrac{3}{2}\)

\(=>x=\dfrac{3}{2}-\dfrac{3}{2}=0\)

25 tháng 3 2022

\(x=\dfrac{4}{27}-\dfrac{2}{3}\)

\(x=-\dfrac{14}{27}\)

25 tháng 3 2022

\(x=\dfrac{4}{27}-\dfrac{2}{3}\)

\(x=-\dfrac{14}{27}\)

5 tháng 3 2017

a) ta co:

1/18<x/12<y/9<1/4

=>2/36<x.3/36<y.4/36<9/36

=>x.3thuộc{3;6};y.4thuộc{4;8}

=>x thuộc{1;2};y thuộc{1:2}

b) ta co

7/8<x/40<9/10

=>70/80<x.2/40<72/80

=>x.2 =71

=>x=71/2

16 tháng 2 2022

\(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)

=>\(B=\dfrac{32}{64}+\dfrac{16}{64}+\dfrac{6}{64}+\dfrac{2}{64}+\dfrac{1}{64}\)

=>\(B=\dfrac{32+16+6+2+1}{64}\)

=>\(B=\dfrac{63}{64}\)

16 tháng 2 2022

\(\dfrac{63}{64}\)

19 tháng 3 2017

 M = 1/21 + 1/28+1/36+...+1/465

     = 2/42+2/56+2/72+...+2/930
     = 2.( 1/6.7 + 1/7.8 + 1/ 7.9 + ... + 1/30.31)

     = 2.( 1/6-1/7+1/7-1/8+...+1/30-1/31)

     = 2.(1/6 - 1/31) = 2.25/186 = 25/92

17 tháng 2 2023

\(\dfrac{3x-4}{4}=\dfrac{4x-8}{5}\)

\(\Leftrightarrow\dfrac{3x-4}{4}-\dfrac{4x-8}{5}=0\)

\(\Leftrightarrow\dfrac{5\left(3x-4\right)}{20}-\dfrac{4\left(4x-8\right)}{20}=0\)

\(\Leftrightarrow\dfrac{15x-20}{20}-\dfrac{16x-32}{20}=0\)

\(\Leftrightarrow\dfrac{15x-20-16x+32}{20}=0\)

\(\Leftrightarrow\dfrac{x-12}{20}=0\)

\(\Leftrightarrow x-12=0\)

\(\Leftrightarrow x=12\)

16 tháng 2 2017

3,375 nhé bạn

16 tháng 2 2017

3.375 bạn nhé

11 tháng 9 2017

a/ \(\dfrac{5}{6}-\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{-5}{12}\)

\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{5}=\dfrac{5}{6}-\dfrac{-5}{12}\)

\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{5}=\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{29}{20}\)

\(\Leftrightarrow x=\dfrac{29}{10}\)

Vậy ...

b/ \(\left(4x-3\right)\left(\dfrac{5}{4}x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-3=0\\\dfrac{5}{4}x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=3\\\dfrac{5}{4}x=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{8}{5}\end{matrix}\right.\)

Vậy .....

c/ \(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|-\dfrac{3}{4}=1,5\)

\(\Leftrightarrow\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=\dfrac{9}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{9}{4}\\\dfrac{7}{8}x-\dfrac{2}{3}=-\dfrac{9}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{8}x=\dfrac{35}{12}\\\dfrac{7}{8}x=-\dfrac{19}{12}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-\dfrac{38}{21}\end{matrix}\right.\)

Vậy ......

d/ \(\left(\dfrac{3}{5}x-\dfrac{1}{2}\right)^3=\dfrac{8}{125}\)

\(\Leftrightarrow\left(\dfrac{3}{5}x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\)

\(\Leftrightarrow\dfrac{3}{5}x-\dfrac{1}{2}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{9}{10}\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy ...

11 tháng 9 2017

a. \(\dfrac{5}{6}-\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{-5}{12}\)

\(\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{5}{6}-\dfrac{-5}{12}\)

\(\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{5}{4}\)

\(\dfrac{3}{6}x=\dfrac{5}{4}+\dfrac{1}{5}\)

\(\dfrac{3}{6}x=\dfrac{29}{20}\)

\(x=\dfrac{29}{20}:\dfrac{3}{6}\)

\(x=\dfrac{29}{10}\)

Vậy...

b. \(\left(4x-3\right).\left(\dfrac{5}{4}x+2\right)=0\)

\(\left[{}\begin{matrix}4x-3=0\\\dfrac{5}{4}x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}4x=3\\\dfrac{5}{4}x=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{-8}{5}\end{matrix}\right.\)

Vậy ...

c. \(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|-\dfrac{3}{4}=1,5\)

\(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=1,5+\dfrac{3}{4}\)

\(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=\dfrac{9}{4}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{9}{4}\\\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{-9}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{8}x=\dfrac{35}{12}\\\dfrac{7}{8}x=\dfrac{-19}{12}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=\dfrac{-38}{21}\end{matrix}\right.\)

Vậy...