Question

\(\dfrac{11}{3}+\left|\dfrac{1}{2}-x\right|=4\)

Answer 1

`11/3+|1/2-x|=4`

`|1/2-x|=4-11/3=1/3`

`@TH1:1/2-x=1/3=>x=1/2-1/3=1/6`

`@TH2:1/2-x=-1/3=>x=1/2-(-1/3)=5/6`

Vậy `x in {1/6;5/6}`

Answer 2

`11/3 + |1/2 -x| = 4`

`|1/2 -x| = 4 - 11/3`

`|1/2-x| = 1/3`

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{5}{6}\end{matrix}\right.\)

Vậy...