Question

giải hệ \(\hept{\begin{cases}x^3-y^3+3x^2+6x-3y+4=0\\x^2+y^2-3x=1\end{cases}}\)

Answer 1

Từ pt (1) ta sẽ có

\(\left(x+1\right)^3-y^3+3\left(x+1-y\right)=0\)

\(\Leftrightarrow\left(x+1-y\right)\left[\left(x+1\right)^2+y\left(x+1\right)+y^2\right]+3\left(x+1-y\right)=0\)

\(\Leftrightarrow\left(x+1-y\right)\left[\left(x+1\right)^2+y\left(x+1\right)+y^2+3\right]=0\)

\(\Leftrightarrow\left(x+1-y\right)\left[\left(x+1+\frac{y}{2}\right)^2+\frac{3y^2}{4}+3\right]=0\)

\(\Leftrightarrow x+1=y\left(Do\left[...\right]>0\right)\)

Thay vô pt (2) ....