\(\sqrt{x^2+2x+2}=x-1\) ĐKXĐ \(x\ge1\)
\(\Leftrightarrow x^2+2x+2=\left(x-1\right)^2\)
\(\Leftrightarrow x^2+2x+2=x^2-2x+1\)
\(\Leftrightarrow4x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{4}\left(loại\right)\)
Vậy \(S=\varnothing\)
Vì \(VT>0\left(x^2+2x+2=\left(x+1\right)^2+1>0\right)\Rightarrow VP>0\Rightarrow x>1\)
\(\sqrt{x^2+2x+2}=x-1\)
\(\Rightarrow x^2+2x+2=\left(x-1\right)^2\Rightarrow x^2+2x+2=x^2-2x+1\)
\(\Rightarrow4x=-1\Rightarrow x=-\dfrac{1}{4}\) (loại)
\(\sqrt{x^2+2x+2}=x-1\\ \Rightarrow x^2+2x+2=\left(x-1\right)^2\\ \Leftrightarrow x^2+2x+2=x^2-2x+1\\ \Leftrightarrow x^2+2x+2-x^2+2x-1=0\\ \Leftrightarrow4x+1=0\\ \Leftrightarrow x=\dfrac{-1}{4}\)
vậy....
