Đáp án:
a, `x=\frac{1}{2}`
b, `x∈{\frac{13}{2};-\frac{11}{2}}`
Giải thích các bước giải:
a, `|x+1|=|x-2|`
`⇔`\(\left[ \begin{array}{l}x+1=x-2\\x+1=2-x\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}0=-3(\text{vô lí})\\x=\dfrac{1}{2}\end{array} \right.\)
`⇔x=\frac{1}{2}`
Vậy `x=\frac{1}{2}`
b, `5-|2x-1|=-7`
`⇔|2x-1|=12`
`⇔`\(\left[ \begin{array}{l}2x-1=12\\2x-1=-12\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{13}{2}\\x=-\dfrac{11}{2}\end{array} \right.\)
Vậy `x∈{\frac{13}{2};-\frac{11}{2}}`