Question

D=-|x+1|+|3-2x| cho|x-2|=1

Answer 1

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

\(D=-\left|x+1\right|+\left|3-2x\right|\)

+) Với x=3

\(D=-\left|3+1\right|+\left|3-2.3\right|=-\left|4\right|+\left|-3\right|=-4+3=-1\)

+) Với x=1

\(D=-\left|x+1\right|+\left|3-2x\right|=-\left|1+1\right|+\left|3-2.1\right|=-2+1=-1\)

Answer 2

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

Với \(x=3\Leftrightarrow D=-\left|3+1\right|+\left|3-6\right|=-4+3=-1\)

Với \(x=1\Leftrightarrow D=-\left|1+1\right|+\left|3-2\right|=-2+1=-1\)

Vậy \(D=-1\)