Question

\(x^2+\left(\frac{x}{x+1}\right)^2=\frac{5}{4}\)

Giải PT trên.

Answer 1

ĐKXĐ: ...

\(\Leftrightarrow x^2+\left(\frac{x}{x+1}\right)^2-\frac{2x^2}{x+1}+\frac{2x^2}{x+1}=\frac{5}{4}\)

\(\Leftrightarrow\left(x-\frac{x}{x+1}\right)^2+\frac{2x^2}{x+1}-\frac{5}{4}=0\)

\(\Leftrightarrow\left(\frac{x^2}{x+1}\right)^2+\frac{2x^2}{x+1}-\frac{5}{4}=0\)

Đặt \(\frac{x^2}{x+1}=t\)

\(t^2+2t-\frac{5}{4}=0\Rightarrow\left[{}\begin{matrix}t=\frac{1}{2}\\t=-\frac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x^2}{x+1}=\frac{1}{2}\\\frac{x^2}{x+1}=-\frac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-x-1=0\\2x^2+5x+5=0\end{matrix}\right.\) (bấm casio)