Đáp án + Giải thích các bước giải:
`a//|x-1|-x+1=0` `(x≥1)`
`⇒|x-1|=x-1`
`⇒` \(\left[ \begin{array}{l}x-1=x-1\\x-1=-x+1\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x-1=x-1\\x+x=1+1\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x-1=x-1\\2x=2\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x-1=x-1\text{(Luôn đúng)}\\x=1(TM)\end{array} \right.\)
Vậy với mọi `x≥1` thì `|x-1|-x+1=0`
`b//|x+7|=|x-9|`
`⇒` \(\left[ \begin{array}{l}x+7=x-9\\x+7=-x+9\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x-x=-7-9\\x+x=-7+9\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}0x=-16\\2x=2\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}0x=-16\text{(Vô lí)}\\x=1(TM)\end{array} \right.\)
Vậy `x=1`
`c//|2-x|+2=x` `(x≥2)`
`⇒|2-x|=x-2`
`⇒` \(\left[ \begin{array}{l}2-x=x-2\\2-x=-x+2\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}-x-x=-2-2\\-x+x=-2+2\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}-2x=-4\\0x=0\text{(Luôn đúng)}\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=2(TM)\\0x=0\text{(Luôn đúng)}\end{array} \right.\)
Vậy với mọi `x≥2` thì `|2-x|+2=x`
