ĐKXĐ: x<>-2
Ta có: \(\frac{x^2}{(x+2)^2}=3(x^2-2x-1)\)
=>\(\left(\frac{x}{x+2}\right)^2=3(x^2-2x-1)\) (1)
Đặt \(y=\frac{x}{x+2}\) (Điều kiện: y<>1)
(1)=> \(y(x+2)=x\)
\(\implies x(1-y)=2y\implies x=\frac{2y}{1-y}\)
\(x^2-2x-1=\left(\frac{2y}{1-y}\right)^2-2\left(\frac{2y}{1-y}\right)-1\)
\(=\frac{4y^2 - 4y(1-y) - (1-y)^2}{(1-y)^2}\)
\(=\frac{7y^2 - 2y - 1}{(1-y)^2}\)
(1) sẽ trở thành: \(y^2=3\cdot\frac{7y^2 - 2y - 1}{(1-y)^2}\)
=>\(y^2(1-y)^2=3(7y^2-2y-1)\)
=>\([y(1-y)]^2=21y^2-6y-3\)
=>\((y-y^2)^2=-21(y-y^2)-3\)
=>\(20y^2 - 6y - 3 = 0\)
=>\(y^2-\frac{3}{10}y-\frac{3}{20}=0\)
=>\(y^2-2\cdot y\cdot\frac{3}{20}+\frac{9}{400}-\frac{9}{400}-\frac{60}{400}=0\)
=>\(\left(y-\frac{3}{20}\right)^2=\frac{69}{400}\)
=>\(\left[\begin{array}{l}y-\frac{3}{20}=\frac{\sqrt{69}}{20}\\ y-\frac{3}{20}=-\frac{\sqrt{69}}{20}\end{array}\right.\Rightarrow\left[\begin{array}{l}y=\frac{3+\sqrt{69}}{20}\\ y=\frac{3-\sqrt{69}}{20}\end{array}\right.\)
\(y = \frac{3 + \sqrt{69}}{20}\)
=>\(x=\frac{2y}{1-y}=\frac{12 + 2\sqrt{69}}{11}\)
\(y=\frac{3-\sqrt{69}}{20}\)
=>\(x=\frac{2y}{1-y}=\frac{12-2\sqrt{69}}{11}\)
