Question

7, Tìm x

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

c) |2x+1|-2x=6                                          d) |x-5|+x=5

Answer 1

Answer 2

a) Ta có: \(\dfrac{1}{4}-\left|x+\dfrac{1}{2}\right|=\dfrac{1}{8}\)

\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{1}{8}\)

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

 

Answer 3

b) Ta có: \(\left|2x-5\right|=x+3\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=\dfrac{2}{3}\left(nhận\right)\end{matrix}\right.\)

Answer 4

c) Ta có: \(\left|2x+1\right|-2x=6\)

\(\Leftrightarrow\left|2x+1\right|=2x+6\)

\(\Leftrightarrow-2x-1=2x+6\left(x< -\dfrac{1}{2}\right)\)

\(\Leftrightarrow-2x-2x=6+1\)

\(\Leftrightarrow-4x=7\)

hay \(x=-\dfrac{7}{4}\left(nhận\right)\)

Answer 5

d) Ta có: \(\left|x-5\right|+x=5\)

\(\Leftrightarrow\left|x-5\right|=5-x\)

\(\Leftrightarrow x-5\le0\)

hay \(x\le5\)