Question

Giải phương trình

\(\frac{1}{x^2+2x-3}+\frac{18}{x^2+2x+2}=\frac{18}{x^2+2x+1}\)

Answer 1

ĐKXĐ : \(x\ne1;-3\)

Đặt \(x^2+2x+1=a\) , ta có :

\(\frac{1}{a-4}+\frac{18}{a+1}=\frac{18}{a}\)

\(\Leftrightarrow\frac{a+1+18a-72}{\left(a+1\right)\left(a-4\right)}=\frac{18}{a}\)

\(\Leftrightarrow\frac{19a-71}{a^2-3a-4}=\frac{18}{a}\)

\(\Leftrightarrow19a^2-71a-18a^2+54a+72=0\)

\(\Leftrightarrow a^2-17a+72=0\)

\(\Leftrightarrow\left(a-8\right)\left(a-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=8\\a=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=8\\\left(x+1\right)^2=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{8}-1\\\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\end{matrix}\right.\)

Vậy ...