Question

a)Rút gọn biểu thức \(B=\sqrt{2\left(\sqrt{x^2+y^2}-x\right)\left(\sqrt{x^2+y^2}-y\right)}+\sqrt{x^2+y^2}\)với x>0,y>0

b) Giải hệ phương trình sau : \(\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^4+x^2y^2+y^4=21\end{matrix}\right.\)

Answer 1

a/

\(B=\sqrt{2\left(x^2+y^2+xy-\left(x+y\right)\sqrt{x^2+y^2}\right)}+\sqrt{x^2+y^2}\)

\(=\sqrt{x^2+y^2+2xy-2\left(x+y\right)\sqrt{x^2+y^2}+x^2+y^2}+\sqrt{x^2+y^2}\)

\(=\sqrt{\left(x+y\right)^2-2\left(x+y\right)\sqrt{x^2+y^2}+\sqrt{x^2+y^2}}+\sqrt{x^2+y^2}\)

\(=\sqrt{\left[x+y-\sqrt{x^2+y^2}\right]^2}+\sqrt{x^2+y^2}\)

\(=x+y-\sqrt{x^2+y^2}+\sqrt{x^2+y^2}=x+y\)

b/

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\x^2+y^2-xy=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+3y^2+3xy=21\\7x^2+7y^2-7xy=21\end{matrix}\right.\)

\(\Rightarrow4x^2+4y^2-10xy=0\)

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

\(\Leftrightarrow...\)