Question

giải hệ phương trình: {xy(x+2)(y+2)=4

và x^2+y^2+2(x+y)=4

Answer 1

\(\left\{{}\begin{matrix}xy\left(x+2\right)\left(y+2\right)=4\\x^2+y^2+2\left(x+y\right)=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\left(x+2\right).y\left(y+2\right)=4\\x\left(x+2\right)+y\left(y+2\right)=4\end{matrix}\right.\)Đặt:\(\left\{{}\begin{matrix}x\left(x+2\right)=a\\y\left(y+2\right)=b\end{matrix}\right.\) thì: \(\left\{{}\begin{matrix}ab=4\\a+b=4\Rightarrow b=4-a\end{matrix}\right.\Rightarrow\left(1\right)\Leftrightarrow a\left(4-a\right)=4\)

\(\Rightarrow4a-a^2-4=0\Leftrightarrow a^2-4a+4=0\Leftrightarrow\left(a-2\right)^2=0\Leftrightarrow a=2\)Tương tự => b=2. Tự tính tiếp nhé