\(\left(x^2+x-1\right)\left(x^2+x+3\right)=5\\ \Leftrightarrow\left(x^2+x-1\right)\left(x^2+x-1+4\right)-5=0\\ \Leftrightarrow\left(x^2+x-1\right)^2+4\left(x^2+x-1\right)-5=0\\ \Leftrightarrow\left[\left(x^2+x-1\right)^2+5\left(x^2+x-1\right)^2\right]-\left[\left(x^2+x-1\right)+5\right]=0\\ \Leftrightarrow\left(x^2+x-1\right)\left(x^2+x-1+5\right)-\left(x^2+x-1+5\right)=0\\ \Leftrightarrow\left(x^2+x-1+5\right)\left(x^2+x-1-1\right)=0\\ \Leftrightarrow\left(x^2+x+4\right)\left(x^2+x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{15}{4}=0\\\left(x^2+2x\right)-\left(x+2\right)=0\end{matrix}\right. \)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\left(vô.lí\right)\\x\left(x+2\right)-\left(x+2\right)=0\end{matrix}\right.\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Đặt x\(^2\) +x+1=a
=>(a-2)(a+2)=5
=>a^2=9
=>a=3
và a=-3
thay ngược vào ta được
1,x^2+x+1=3
<=>x^2+x-2=0
<=>(x-1)(x+2)=0
<=>x=1 hoặc x=-2
2,x^2+x+1=-3
<=>x^2+x+4=0
<=>(x+\(\dfrac{1}{2}\) )^2+\(\dfrac{15}{4}\) =0 (vô nghiệm)
Vậy tập nghiệm S=(1;-2)
