\(x^2+\frac{4x^2}{\left(x+2\right)^2}=12\left(ĐK:x\ge-2\right)\)
Thêm \(-2x\cdot\frac{2x}{x+2}\) vào hai vế ta được:
\(x^2-2x\cdot\frac{2x}{x+2}+\frac{4x^2}{\left(x+2\right)^2}=12-2x\cdot\frac{4x^2}{x+2}\)
\(\Leftrightarrow\left(x-\frac{2x}{x+2}\right)^2=12-\frac{4x^2}{x+2}\)
\(\Leftrightarrow\left(\frac{x^2}{x+2}\right)^2+\frac{4x^2}{x+2}-12=0\)
Đặt: \(\frac{x^2}{x+2}=a\), khi đó pt trở thành:
\(a^2+4a-12=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+6\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}a-2=0\\a+6=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}a=2\\a=-6\end{array}\right.\)
Với a=2 ta có:\(\frac{x^2}{x+2}=2\)
\(\Leftrightarrow x^2=2x+4\)
\(\Leftrightarrow x^2-2x+1=5\)
\(\Leftrightarrow\left(x-1\right)^2=5\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=\sqrt{5}\\x-1=-\sqrt{5}\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1+\sqrt{5}\left(tm\right)\\x=1-\sqrt{5}\left(tm\right)\end{array}\right.\)
Với a=-6 ta có: \(\frac{x^2}{x+2}=-6\)
\(\Leftrightarrow x^2=-6x-12\)
\(\Leftrightarrow x^2+6x+12=0\)
\(\Leftrightarrow\left(x^2+3\right)^2+3=0\) ( vô nghiệm)
Vậy pt đã cho có tập nghiệm là \(S=\left\{1-\sqrt{5};1+\sqrt{5}\right\}\)
