ĐK : \(x\in R\)
Đặt \(x^2+x=a\) . Phương trình trở thành :
\(a-\dfrac{7}{a+1}=5\)
\(\Leftrightarrow\dfrac{a\left(a+1\right)}{a+1}-\dfrac{7}{a+1}=\dfrac{5\left(a+1\right)}{a+1}\)
\(\Leftrightarrow a^2+a-7=5a+5\)
\(\Leftrightarrow a^2-4a-12=0\)
\(\Delta=\left(-4\right)^2+4.12=16+48=64>0\)
\(\Rightarrow\left\{{}\begin{matrix}a_1=\dfrac{4+\sqrt{64}}{2}=6\\a_2=\dfrac{4-\sqrt{64}}{2}=-2\end{matrix}\right.\)
Với : \(a=6\)
\(\Leftrightarrow x^2+x=6\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Delta=1+4.6=25>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-1+\sqrt{25}}{2}=2\\x_2=\dfrac{-1-\sqrt{25}}{2}=-3\end{matrix}\right.\)
Với \(a=-2\)
\(\Leftrightarrow x^2+x=-2\)
\(\Leftrightarrow x^2+x+2=0\)
\(\Delta=1-4.2=-7< 0\)
Nên phương trình vô nghiệm .
Vậy \(S=\left\{-3;2\right\}\)
