a/ ĐKXĐ: ...
Đặt \(\sqrt{2x-2}=2t\ge0\Rightarrow x=2t^2+1\)
Pt trở thành:
\(2t\left(2t^2+1\right)+5\left(2t^2+1\right)=9\)
\(\Leftrightarrow2t^3+5t^2+t-2=0\)
\(\Leftrightarrow\left(2t-1\right)\left(t+1\right)\left(t+2\right)=0\)
\(\Leftrightarrow t=\frac{1}{2}\Rightarrow\sqrt{2x-2}=1\)
\(\Rightarrow x=\frac{3}{2}\)
b/ ĐKXĐ: ...
Đặt \(\sqrt{x+2}=t-1\ge0\)
\(\Rightarrow x+2=\left(t-1\right)^2\Rightarrow x=t^2-2t-1\)
Ta được hệ: \(\left\{{}\begin{matrix}t-1=x^2-2x-2\\x=t^2-2t-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t=x^2-2x-1\\x=t^2-2t-1\end{matrix}\right.\)
\(\Rightarrow x^2-t^2-2x+2t=t-x\)
\(\Leftrightarrow\left(x-t\right)\left(x+t\right)-\left(x-t\right)=0\)
\(\Leftrightarrow\left(x-t\right)\left(x+t-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=x\\t-1=-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}=x-1\left(x\ge1\right)\\\sqrt{x+2}=-x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x^2-2x+1\\x+2=x^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x-1=0\\x^2-x-2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{3+\sqrt{13}}{2}\\x=-1\end{matrix}\right.\)
