Question

\(\hept{\begin{cases}^{x^2+y+x^3y+xy^2+xy+\frac{5}{4}=0}\\x^4+y^2+xy\left(1+2x\right)+\frac{5}{4}=0\end{cases}}\)

Answer 1

Lấy \(PT\left(2\right)-PT\left(1\right)\) ta được :

\(x^4+y^2+2x^2y-x^2-y-x^3y-xy^2=0\)

\(\Leftrightarrow\left(x^2+y\right)^2-\left(x^2+y\right)-xy\left(x^2+y\right)=0\)

\(\Leftrightarrow\left(x^2+y\right)\left(x^2+y-xy-1\right)=0\)

\(\Leftrightarrow\left(x^2+y\right)\left[\left(x-1\right)\left(x+1\right)-y\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x^2+y\right)\left(x-y+1\right)\left(x-1\right)=0\)

Xét các TH xong thay vô

Answer 2

Xét sao vậy?