ĐKXĐ: ...
a/ \(x^2+\frac{4x^2}{\left(x+2\right)^2}-\frac{4x^2}{x+2}+\frac{4x^2}{x+2}-12=0\)
\(\Leftrightarrow\left(x-\frac{2x}{x+2}\right)^2+\frac{4x^2}{x+2}-12=0\)
\(\Leftrightarrow\left(\frac{x^2}{x+2}\right)^2+\frac{4x^2}{x+2}-12=0\)
Đặt \(\frac{x^2}{x+2}=t\Rightarrow t^2+4t-12=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{x^2}{x+2}=2\\\frac{x^2}{x+2}=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+6x+12=0\end{matrix}\right.\)
b/ Giống câu a:
\(x^2+\left(\frac{x}{x+1}\right)^2-\frac{2x^2}{x+1}+\frac{2x^2}{x+1}-\frac{5}{4}=0\)
\(\Leftrightarrow\left(x-\frac{x}{x+1}\right)^2+\frac{2x^2}{x+1}-\frac{5}{4}=0\)
\(\Leftrightarrow\left(\frac{x^2}{x+1}\right)^2+\frac{2x^2}{x+1}-\frac{5}{4}=0\)
\(\Leftrightarrow t^2+2t-\frac{5}{4}=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{1}{2}\\t=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{x^2}{x+1}=\frac{1}{2}\\\frac{x^2}{x+1}=-\frac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-x-1=0\\2x^2+5x+5=0\end{matrix}\right.\)
c/ Đặt \(x^2-2x+2=t\ge1\)
\(\frac{1}{t}+\frac{2}{t+1}=\frac{6}{t+2}\)
\(\Leftrightarrow\frac{3t+1}{t^2+t}=\frac{6}{t+2}\)
\(\Leftrightarrow\left(3t+1\right)\left(t+2\right)=6\left(t^2+t\right)\)
\(\Leftrightarrow3t^2-t-2=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{2}{3}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^2-2x+2=1\Rightarrow x=1\)