Question

Tìm x: | 2x + 3/4|= 1/2 

 

Answer 1

\(\left|2x+\dfrac{3}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{3}{4}=\dfrac{1}{2}\\2x+\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{4}\\2x=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{8}\\x=-\dfrac{5}{8}\end{matrix}\right.\)

Answer 2

\(\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\left[{}\begin{matrix}2x=\dfrac{5}{4}\\2x=\dfrac{1}{4}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)

Answer 3

\(\left|2x+\dfrac{3}{4}\right|=\dfrac{1}{2}\)

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

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

\(\Rightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{4}\\2x=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}:2=-\dfrac{1}{8}\\x=-\dfrac{5}{4}:2=-\dfrac{5}{8}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{-1}{8};-\dfrac{5}{8}\right\}\)

Hoctot