Bài 5:
\(\left\{{}\begin{matrix}x^2-x-y=-1\left(1\right)\\x^3+x^2y+x^2+xy=x+y-2\left(2\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow x+y=x^2+1\)
Ta có:
\(\left(2\right)\Leftrightarrow x^2\left(x+y\right)+x\left(x+y\right)=x+y-2\)
Thay `x+y=x^2+1` vào (2) ta có:
\(x^2\left(x^2+1\right)+x\left(x^2+1\right)=\left(x^2+1\right)-2\\ \Leftrightarrow x^4+x^2+x^3+x=x^2-1\\ \Leftrightarrow x^4+x^3+x+1=0\\ \Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^3+1\right)\\ \Leftrightarrow\left(x+1\right)^2\left(x^2-x+1\right)=0\)
Mà: `x^2-x+1≠0=>(x+1)^2=0<=>x=-1`
Thay `x=-1` vào (1) ta có:
\(\left(-1\right)+y=\left(-1\right)^2+1\\ \Leftrightarrow y-1=2\\ \Leftrightarrow y=1+2=3\)
Vậy nghiệm của hpt là: `(-1;3)`
