Đặt 2x2+3x-2=a,x2-1=b => x2+3x=a-b+1
Pt tương đương
\(\frac{3x+1}{a}+\frac{1}{b}=\frac{1}{a-b+1}\)
\(\frac{3xb+a+b}{ab}=\frac{1}{a-b+1}\)
=>(3xb+a+b)(a-b+1)=ab
=>3xab+a2-3xb2-ab-b2+3xb+a+b=0
Đến đây bạn tự giải tiếp nhé
\(\frac{3x+1}{2x^2+3x-2}+\frac{1}{x^2-1}=\frac{1}{x^2+3x}\left(1\right)\)
ĐKXĐ: \(2x^2+3x-2=\left(x-2\right)\left(2x-1\right)\ne0\)
\(x^2-1=\left(x-1\right)\left(x+1\right)\ne0\)
\(x^2+3x=x\left(x+3\right)\ne0\)
\(\Rightarrow x\notin\left\{2;\frac{1}{2};1;-1;0;-3\right\}\)
Ta có: \(\left(1\right)\Leftrightarrow\frac{3x+1}{2x^2+3x-2}+\frac{1}{x^2-1}-\frac{1}{x^2+3x}=0\)
\(\Leftrightarrow\frac{3x+1}{2x^2+3x-2}+\frac{3x+1}{\left(x^2-1\right)\left(x^2+3x\right)}=0\)
\(\Leftrightarrow\left(3x+1\right)\left(\frac{1}{2x^2+3x-2}+\frac{1}{\left(x^2-1\right)\left(x^2+3x\right)}\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(\frac{\left(x^2-1\right)\left(x^2+3x\right)+\left(2x^2+3x-2\right)}{\left(x^2-1\right)\left(x^2+3x\right)\left(2x^2+3x-2\right)}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\2x^2+3x-2=-\left(x^4+3x^3-x^2-3x^2\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x^4+3x^3+x^2-2=0\left(2\right)\end{cases}}\)
Ta có: \(\left(2\right)\Leftrightarrow\left(x^2+2x-2\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow x^2+2x-2=0\)
(vì \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Leftrightarrow\orbr{\begin{cases}x1=-1-\sqrt{3}\\x2=-1+\sqrt{3}\end{cases}}\)
Vậy \(S=\left\{\frac{-1}{3};-1-\sqrt{3};-1+\sqrt{3}\right\}\)
