Đáp án:
`x\in {-10;-20/3;6;28/3}.`
Giải thích các bước giải:
`||x+1/3|-8|=5/3`
TH1:
`|x+1/3|-8=5/3`
`<=>|x+1/3|=8+5/3`
`<=>|x+1/3|=29/3`
\(⇔\left[ \begin{array}{l}x+\dfrac13=\dfrac{29}{3}\\x+\dfrac13=-\dfrac{29}{3}\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\dfrac{28}{3}\\x=-\dfrac{30}{3}=-10\end{array} \right.\)
TH2"
`|x+1/3|-8=-5/3`
`<=>|x+1/3|=19/3`
\(⇔\left[ \begin{array}{l}x+\dfrac13=\dfrac{19}{3}\\x+\dfrac13=-\dfrac{19}{3}\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\dfrac{18}{3}=6\\x=-\dfrac{20}{3}\end{array} \right.\)
Vậy `x\in {-10;-20/3;6;28/3}.`
