Question

\(Ghpt:\left\{{}\begin{matrix}x^2+y^2+\dfrac{8xy}{x+y}=16\\\sqrt{x+y}=x^2-y\end{matrix}\right.\)

Answer 1

\(x^2+y^2+2xy-16-2xy+\dfrac{8xy}{x+y}=0\)

\(\Leftrightarrow\left(x+y\right)^2-16-2xy\left(1-\dfrac{4}{x+y}\right)=0\)

\(\Leftrightarrow\left(x+y-4\right)\left(x+y+4\right)-2xy\left(\dfrac{x+y-4}{x+y}\right)=0\)

\(\Leftrightarrow\left(x+y-4\right)\left(x+y+4-\dfrac{2xy}{x+y}\right)=0\)

\(\Rightarrow\left(x+y-4\right)\left(x^2+y^2+4x+4y\right)=0\)

\(\Rightarrow x+y=4\) (do \(x+y>0\) theo ĐKXĐ nên \(x^2+y^2+4\left(x+y\right)>0\))

Rồi thế vào pt dưới