Question

Giải hệ phương trình

\(\left\{{}\begin{matrix}\left(x^2+y^2\right)\left(x^2-y^2\right)=144\\\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=y\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}\left(x^2+y^2\right)\left(x^2-y^2\right)=144\left(1\right)\\\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=y\left(2\right)\end{matrix}\right.\)

\(\Rightarrow\left(2\right)\Leftrightarrow2x^2-2\sqrt{\left(x^2+y^2\right)\left(x^2-y^2\right)}=y^2\)

\(\Leftrightarrow y^2=2x^2-24\)Thế vô (1) ta được

\(\left(x^2+2x^2-24\right)\left(x^2-2x^2+24\right)=144\)

\(\Leftrightarrow x^4-32x^2+240=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm2\sqrt{5}\\x=\pm2\sqrt{3}\end{matrix}\right.\)

Làm tiếp nhé