\(\left(x-2\right)\left(x^2+6x-11\right)^2=\left(5x^2-10x+1\right)^2\) \(\Rightarrow x>2\)
\(\Rightarrow x^2+6x-11>0\)
\(pt\Leftrightarrow x-2=\left(\dfrac{5x^2-10x+1}{x^2+6x-11}\right)^2\Leftrightarrow\sqrt{x-2}=\dfrac{5x^2-10x+1}{x^2+6x-11}\)
\(\Leftrightarrow\sqrt{x-2}-1=\dfrac{5x^2-10x+1}{x^2+6x-11}-1=\dfrac{4x^2-16x+12}{x^2+6x+12}\)
\(\Leftrightarrow\dfrac{x-3}{\sqrt{x-2}+1}=\dfrac{4\left(x-1\right)\left(x-3\right)}{x^2+6x-11}\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\\dfrac{1}{\sqrt{x-2}+1}=\dfrac{4\left(x-1\right)}{x^2+6x-11}\left(1\right)\end{matrix}\right.\)
Xét (1):
\(x^2+6x-11=4\left(x-1\right)+4\left(x-1\right)\sqrt{x-2}\)
\(\Leftrightarrow x^2+2x-7-4\left(x-1\right)\sqrt{x-2}=0\)
\(\Leftrightarrow x^2-2x+1-2\left(x-1\right)\sqrt{4x-8}+4x-8=0\)
\(\Leftrightarrow\left(x-1\right)^2-2\left(x-1\right)\sqrt{4x-8}+\left(\sqrt{4x-8}\right)^2=0\)
\(\Leftrightarrow\left(x-1-\sqrt{4x-8}\right)^2=0\)
\(\Leftrightarrow x-1=\sqrt{4x-8}\)
\(\Leftrightarrow x^2-2x+1=4x-8\)
\(\Leftrightarrow\left(x-3\right)^2=0\Rightarrow x=3\)
Vậy pt đã cho có nghiệm duy nhất \(x=3\)
Đặt \(y=x-2\), phương trình đã cho trở thành:
\( y{\left[ {{{\left( {y + 2} \right)}^2} + 6\left( {y + 2} \right) - 11} \right]^2} = {\left[ {5{{\left( {y + 2} \right)}^2} - 10\left( {y + 2} \right) + 1} \right]^2}\\ \Leftrightarrow y{\left( {{y^2} + 10y + 5} \right)^2} = {\left( {5{y^2} + 10y + 1} \right)^2}\\ \Leftrightarrow {y^5} - 5{y^4} + 10{y^3} - 10{y^2} + 5y - 1 = 0 \Leftrightarrow {\left( {y - 1} \right)^5} = 0 \Leftrightarrow y = 1 \)
Với \(y=1\) ta có \(x-2=1\) \(\Rightarrow x=3\)
Vậy \(x = 3 \)