Question

\(\frac{2}{x^2+x+1}+\frac{3}{x^2+x+2}-\frac{2}{x\left(x+1\right)}=0\)

Giải phương trình trên

Answer 1

ĐKXĐ:x\(\ne\) 0 ; x\(\ne\) -1.

Đặt : x² +x=t ta có pt:

\(\frac{2}{t+1}+\frac{3}{t+2}-\frac{2}{t}\) =0

\(\Leftrightarrow\) \(\frac{2\left(t+2\right)t+3\left(t+1\right)t-2\left(t+1\right)\left(t+2\right)}{\left(t+2\right)\left(t+1\right)t}\)

\(\Rightarrow\) 2t²+4t+3t²+3t-2t²-6t-4=0

\(\Leftrightarrow\) 3t²+t-4=0

\(\Leftrightarrow\) (t-1)(4t+3)=0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}t=1\\t=\frac{-3}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-1=0\\x^2+x+\frac{3}{4}=0\end{matrix}\right.\) \(\Leftrightarrow x^2+x-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{-1-\sqrt{5}}{2}\\\frac{-1+\sqrt{5}}{2}\end{matrix}\right.\)

Vậy : ....................................