ĐKXĐ: \(x\ne-1\)
\(x^2+\dfrac{x^2}{\left(x+1\right)^2}-\dfrac{x^2}{x+1}+\dfrac{x^2}{x+1}=3\)
\(\Leftrightarrow\left(x-\dfrac{x}{x+1}\right)^2+\dfrac{x^2}{x+1}=3\)
\(\Leftrightarrow\left(\dfrac{x^2}{x+1}\right)^2+\dfrac{x^2}{x+1}-3=0\)
Đặt \(\dfrac{x^2}{x+1}=t\Rightarrow t^2+t-3=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{-1+\sqrt{13}}{2}\\t=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x^2}{x+1}=\dfrac{-1+\sqrt{13}}{2}\\\dfrac{x^2}{x+1}=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-\left(\dfrac{-1+\sqrt{13}}{2}\right)x-\dfrac{-1+\sqrt{13}}{2}=0\\x^2+\left(\dfrac{1+\sqrt{13}}{2}\right)x+\dfrac{1+\sqrt{13}}{2}=0\end{matrix}\right.\)
Nghiệm xấu vậy, vế phải là số 2 thì hợp lý hơn