Question

\(a)\frac{2}{5}:(2x+\frac{3}{4})=\frac{-7}{10}\)

\(b)\frac{x+1}{3}=\frac{2-x}{2}\)

\(c)|x-\frac{3}{5}|.\frac{1}{2}-\frac{1}{5}=0\)

\(^{d)x^2-4x=0}\)

Tìm x giúp mình nha!!!

Answer 1

a) \(\frac{2}{5}:\left(2x+\frac{3}{4}\right)=-\frac{7}{10}\)

=> \(2x+\frac{3}{4}=-\frac{7}{10}:\frac{2}{5}\)

=> \(2x+\frac{3}{4}=-\frac{7}{4}\)

=> \(2x=\frac{-7}{4}-\frac{3}{4}\)

=> \(2x=-\frac{5}{2}\)

=> \(x=\frac{-5}{2}:2\)

=> \(x=\frac{-5}{4}\)

b) \(\frac{x+1}{3}=\frac{2-x}{2}\)

\(\Rightarrow2\left(x+1\right)=3\left(2-x\right)\)

\(\Rightarrow2x+2=6-3x\)

\(\Rightarrow2x-3x=6-2\)

\(\Rightarrow-x=4\)

\(\Rightarrow x=4\)

Answer 2

c) \(\left|x-\frac{3}{5}\right|.\frac{1}{2}-\frac{1}{5}=0\)

\(\Rightarrow\left|x-\frac{3}{5}\right|.\frac{1}{2}=\frac{1}{5}\)

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

\(\Rightarrow\left|x-\frac{3}{5}\right|=\frac{2}{5}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}=\frac{2}{5}\\x-\frac{3}{5}=-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}+\frac{2}{5}\\x=\frac{3}{5}+-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)

d) \(x^2-4x=0\)

Ta có : \(x^2-4x=0\)

\(\Rightarrow xx-4x=0\)

\(\Rightarrow x\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=0+4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

Answer 3

\(\text{d) }x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)