Question

Nhờ giải dùm

Answer 1

ĐK: vì VT\(\ge0\) nên VP\(\ge0\Rightarrow x+1\ge0\Rightarrow x\ge-1\)

\(\sqrt{x^2-4x+4}=x+1\Leftrightarrow\sqrt{\left(x-2\right)^2}=x+1\Leftrightarrow\left|x-2\right|=x+1\)Nếu \(-1\le x< 2\) thì

\(\left|x-2\right|=x+1\Leftrightarrow2-x=x+1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

Nếu \(x\ge2\) thì

\(\left|x-2\right|=x+1\Leftrightarrow x-2=x+1\Leftrightarrow0=3\left(ktm\right)\)

Vậy S={\(\dfrac{1}{2}\)}