Question

Giải hệ phương trình :

\(\left\{{}\begin{matrix}x^2-y^2+2y=1\\\left(x+y\right)^2-2x-2y=0\end{matrix}\right.\)

Answer 1

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

Answer 2

\(\left\{{}\begin{matrix}x^2-y^2+2y=1\left(1\right)\\\left(x+y\right)^2-2x-2y=0\left(2\right)\end{matrix}\right.\)

(1)<=>(x-y)(x+y)+2y=1(3)

(2)<=>(x+y)(x+y-2)=0<=>\(\left[{}\begin{matrix}x+y=0\\x+y-2=0\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}x+y=0\left(a\right)\\x+y=2\left(b\right)\end{matrix}\right.\)

thay (a) vào (3): 2y=1<=>y=\(\dfrac{1}{2}\)=>x=\(\dfrac{-1}{2}\)

thay (b) vào (3): 2(x-y)+2y=1<=>2x=1<=>x=\(\dfrac{1}{2}\)=>y=2-\(\dfrac{1}{2}\)=\(\dfrac{3}{2}\)