ĐKXĐ: \(x\ne-1\)
\(\left(\dfrac{x^2-5x}{x+1}\right)\left(\dfrac{x^2+5}{x+1}\right)=\dfrac{-21}{4}\)
Đặt \(\dfrac{x^2+5}{x+1}=a\Rightarrow\dfrac{x^2-5x}{x+1}=a-5\)
Phương trình đã cho trở thành:
\(\left(a-5\right).a=\dfrac{-21}{4}\Leftrightarrow a^2-5a+\dfrac{21}{4}=0\) \(\Rightarrow\left[{}\begin{matrix}a=\dfrac{7}{2}\\a=\dfrac{3}{2}\end{matrix}\right.\)
TH1: \(\dfrac{x^2+5}{x+1}=\dfrac{3}{2}\Leftrightarrow2x^2-3x+7=0\) \(\Rightarrow\) vô nghiệm
TH2: \(\dfrac{x^2+5}{x+1}=\dfrac{7}{2}\Leftrightarrow2x^2-7x+3=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy pt đã cho có 2 nghiệm: \(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
