1) Ta có: \(2\cdot\left|\frac{1}{2}x-\frac{3}{8}\right|-\frac{3}{2}=\frac{1}{4}\)

⇔\(2\cdot\left|\frac{1}{2}x-\frac{3}{8}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)

⇔\(\left|\frac{1}{2}x-\frac{3}{8}\right|=\frac{7}{4}:2=\frac{7}{4}\cdot\frac{1}{2}=\frac{7}{8}\)

⇔\(\left[{}\begin{matrix}\frac{1}{2}x-\frac{3}{8}=\frac{7}{8}\\\frac{1}{2}x-\frac{3}{8}=\frac{-7}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x=\frac{10}{8}\\\frac{1}{2}x=\frac{-4}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{10}{8}:\frac{1}{2}=\frac{10}{8}\cdot2=\frac{20}{8}=\frac{5}{2}\\x=\frac{-4}{8}:\frac{1}{2}=-\frac{4}{8}\cdot2=-\frac{8}{8}=-1\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{5}{2};-1\right\}\)

2) Ta có: \(-5\cdot\left(x+\frac{1}{5}\right)-\frac{1}{2}\cdot\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)

⇔\(-5x-1-\frac{1}{2}x+\frac{1}{3}-\frac{3}{2}x+\frac{5}{6}=0\)

\(\Leftrightarrow-7x+\frac{1}{6}=0\)

\(\Leftrightarrow-7x=-\frac{1}{6}\)

hay \(x=\frac{1}{42}\)

Vậy: \(x=\frac{1}{42}\)

3) Ta có: \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)

\(\Leftrightarrow3x-\frac{3}{2}-5x-3+x-\frac{1}{5}=0\)

\(\Leftrightarrow-x-\frac{47}{10}=0\)

⇔\(-x=\frac{47}{10}\)

hay \(x=\frac{-47}{10}\)

Vậy: \(x=\frac{-47}{10}\)

4) Ta có: \(\frac{3}{4}-2\left|2x-0,125\right|=2\)

\(\Leftrightarrow2\left|2x-\frac{1}{8}\right|=\frac{3}{4}-2=-\frac{5}{4}\)

⇔\(\left|2x-\frac{1}{8}\right|=-\frac{5}{8}\)(vô lý)

Vậy: x∈∅

5) Ta có: \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)

⇔\(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\frac{-7}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x=\frac{7}{8}+\frac{1}{3}=\frac{29}{24}\\\frac{1}{2}x=-\frac{7}{8}+\frac{1}{3}=-\frac{13}{24}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{24}:\frac{1}{2}=\frac{29}{24}\cdot2=\frac{29}{12}\\x=-\frac{13}{24}:\frac{1}{2}=-\frac{13}{24}\cdot2=-\frac{13}{12}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)

Bài mk sai r nhé!!

\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{4}{14}-\frac{2}{13}}\times\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

\(=\frac{\frac{2}{6}+\frac{2}{14}-\frac{2}{26}}{\frac{4}{6}+\frac{4}{14}-\frac{4}{26}}\times\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{356}}{\frac{4}{4}-\frac{4}{16}+\frac{4}{64}-\frac{4}{256}}+\frac{5}{8}\)

\(=\frac{2\left(\frac{1}{6}+\frac{1}{14}-\frac{1}{26}\right)}{4\left(\frac{1}{6}+\frac{1}{14}-\frac{1}{26}\right)}\times\frac{3\left(\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\frac{1}{356}\right)}{4\left(\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)

\(=\frac{2}{4}\times\frac{3}{4}+\frac{5}{8}\)

\(=\frac{1}{2}\times\frac{3}{4}+\frac{5}{8}\)

\(=\frac{3}{8}+\frac{5}{8}\)

\(=\frac{8}{8}=1\)

\(\frac{\frac{109}{3.7.13}}{\frac{361}{3.14.13}}\)\(\frac{\frac{153}{256}}{\frac{51}{64}}\)+5/8

=\(\frac{327}{722}\)+5/8

=\(\frac{3113}{2888}\)

B=\(6\frac{4}{9}-4\frac{4}{9}+3\frac{7}{11}\)

B=\(2+3\frac{7}{11}\)

B=\(5\frac{7}{11}\)

B = \(5\frac{7}{11}=\frac{62}{11}\)

C = 1

D = \(\frac{5}{2}=2\frac{1}{2}\)

a) Ta có: \(\frac{x+1}{3}=\frac{2}{6}\)

⇔\(x=\frac{2\cdot3}{6}-1=\frac{6}{6}-1=1-1=0\)

Vậy: x=0

b) Ta có: \(\frac{x-1}{4}=\frac{1}{-2}\)

⇔\(x=\frac{1\cdot4}{-2}+1=\frac{4}{-2}+1=-1\)

Vậy: x=-1

c) Ta có: \(\frac{-1}{6}=\frac{3}{2x}\)

⇔\(2x=\frac{3\cdot6}{-1}=-18\)

hay x=-9

Vậy: x=-9

d) Ta có: \(\frac{x+1}{3}=\frac{3}{x+1}\)

⇔\(\left(x+1\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy: x∈{2;-4}

e) Ta có: \(\frac{4}{5}=\frac{-12}{9-x}\)

⇔\(9-x=\frac{-12\cdot5}{4}=-15\)

hay x=24

Vậy: x=24

f) Ta có: \(\frac{x-1}{-4}=\frac{-4}{x-1}\)

⇔\(\left(x-1\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy: x∈{5;-3}

g) Ta có: \(\frac{5-x}{2}=\frac{2}{5-x}\)

⇔\(\left(5-x\right)^2=4\)

⇔\(\left[{}\begin{matrix}5-x=2\\5-x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)

Vậy: x∈{3;7}

h) Ta có: \(\frac{4-x}{-5}=\frac{-5}{4-x}\)

⇔\(\left(4-x\right)^2=25\)

⇔\(\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)

Vậy: x∈{-1;9}

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