a/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow\sqrt{x+1}=1+\sqrt{x-2}\)
\(\Leftrightarrow x+1=1+x-2+2\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{x-2}=1\)
\(\Leftrightarrow x=3\)
b/ ĐKXĐ: \(x^2\ge2\)
Đặt \(\sqrt{x^2-2}=t\ge0\Rightarrow x^2=t^2+2\)
Pt trở thành: \(t^2+2-t=4\)
\(\Leftrightarrow t^2-t-2=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=2\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-2}=2\Leftrightarrow x^2=6\Rightarrow x=\pm\sqrt{6}\)
c/
\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=5-\left(x+1\right)^2\)
Do \(\left(x+1\right)^2\ge0\) ;\(\forall x\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{3\left(x+1\right)^2+4}\ge\sqrt{0+4}=2\\\sqrt{5\left(x+1\right)^2+9}\ge\sqrt{0+9}=3\end{matrix}\right.\)
\(\Rightarrow VT\ge5\)
\(VP=5-\left(x+1\right)^2\le5\)
\(\Rightarrow VT\ge VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\left(x+1\right)^2=0\Leftrightarrow x=-1\)
