ĐKXĐ: x>=1
\(2\left(\sqrt{x-1}+1\right)=x+\sqrt{x+2}\)
=>\(2\sqrt{x-1}+2=x+\sqrt{x+2}\)
=>\(\sqrt{4x-4}-\sqrt{x+2}-x+2=0\)
=>\(\dfrac{4x-4-x-2}{\sqrt{4x-4}+\sqrt{x+2}}-\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(\dfrac{3}{\sqrt{4x-4}+\sqrt{x+2}}-1\right)=0\)
=>x-2=0
=>x=2
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{4x-4}+2=x+\sqrt{x+2}\)
\(\Leftrightarrow\sqrt{4x-4}=x-2+\sqrt{x+2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{4x-4}=a\ge0\\\sqrt{x+2}=b>0\end{matrix}\right.\)
\(\Rightarrow a^2-b^2=3x-6=3\left(x-2\right)\)
\(\Rightarrow x-2=\dfrac{a^2-b^2}{3}\)
Pt trở thành:
\(a=\dfrac{a^2-b^2}{3}+b\)
\(\Leftrightarrow a^2-b^2-3\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)-3\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=3-b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4x-4}=\sqrt{x+2}\left(1\right)\\\sqrt{4x-4}=3-\sqrt{x+2}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4x-4=x+2\Rightarrow x=2\)
Xét (2):
\(\Rightarrow4x-4=9-6\sqrt{x+2}+x+2\)
\(\Leftrightarrow2\sqrt{x+2}=5-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le5\\4\left(x+2\right)=x^2-10x+25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le5\\x^2-14x+17=0\end{matrix}\right.\)
\(\Rightarrow x=7-4\sqrt{2}\)
Thay vào (2) thỏa mãn
Vậy pt có 2 nghiệm: \(\left[{}\begin{matrix}x=2\\x=7-4\sqrt{2}\end{matrix}\right.\)
