\(x^2+2\left(2+\sqrt{x-1}\right)=5x\)
\(\Leftrightarrow x^2+4+2\sqrt{x-1}-5x=0\)
\(\Leftrightarrow x^2-5x+2\sqrt{x-1}+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(x^2+2\left(2+\sqrt{x-1}\right)=5x\left(1\right)\)
Đk: \(x\ge1\)
\(\left(1\right)\Leftrightarrow x^2-5x+4+2\sqrt{x-1}=0\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left[\left(x-1\right)-2\sqrt{x-1}+1\right]=0\)
\(\Leftrightarrow\left(x-2\right)^2-\left(\sqrt{x-1}-1\right)^2=0\)
\(\Leftrightarrow\left(x+\sqrt{x-1}-3\right)\left(x-\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{x-1}-3=0\left(2\right)\\x-\sqrt{x-1}-1=0\left(3\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left(x-1\right)+\sqrt{x-1}-2=0\)
\(\Leftrightarrow\left(x-1\right)-\sqrt{x-1}+2\sqrt{x-1}-2=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-1\right)+2\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-1}+2\right)=0\)
\(\Leftrightarrow\sqrt{x-1}-1=0\) (vì \(\sqrt{x-1}+2>0\))
\(\Leftrightarrow x=2\left(nhận\right)\)
\(\left(3\right)\Leftrightarrow\left(x-1\right)-\sqrt{x-1}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x-1}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\left(nhận\right)\)
Vậy phương trình (1) có 2 nghiệm là \(x=1\text{v}àx=2\)
