d)ĐK:\(x,y\ge2\)

Trừ từng vế 2 pt ta được:

\(\sqrt{x+5}-\sqrt{y+5}=\sqrt{x-2}-\sqrt{y-2}\)

\(\Leftrightarrow\frac{x-y}{\sqrt{x+5}+\sqrt{y+5}}=\frac{x-y}{\sqrt{x-2}+\sqrt{y-2}}\)

\(\Leftrightarrow\left(x-y\right)\left(\frac{1}{\sqrt{x+5}+\sqrt{y+5}}-\frac{1}{\sqrt{x-2}+\sqrt{y-2}}\right)=0\)

Do \(\frac{1}{\sqrt{x+5}+\sqrt{y+5}}-\frac{1}{\sqrt{x-2}+\sqrt{y-2}}=0\) vô nghiệm nên \(x=y\)

Khi đó hệ trở thành \(\left\{\begin{matrix}x=y\\\sqrt{x+5}+\sqrt{x-2}=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{\begin{matrix}x=y\\2x+3+2\sqrt{\left(x+5\right)\left(x-2\right)}=49\end{matrix}\right.\)

\(\Leftrightarrow\left\{\begin{matrix}x=y\\\sqrt{\left(x+5\right)\left(x-2\right)}=23-x\end{matrix}\right.\)

\(\Leftrightarrow\left\{\begin{matrix}x=y\\x\le23\\49x=539\end{matrix}\right.\)\(\Leftrightarrow x=y=11\) (thỏa mãn)

VP của câu c sai

c)Ta thấy x=0 ko là nghiệm của hệ

Chia 2 vế cho x² ta có: \(hpt\Leftrightarrow\left\{\begin{matrix}\frac{y}{x^2}+\frac{y^2}{x}=6\\\frac{1}{x^2}+y^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{\begin{matrix}\frac{y}{x}\left(\frac{1}{x}+y\right)=6\\\left(\frac{1}{x}+y\right)^2-2\frac{y}{x}=5\end{matrix}\right.\).Đặt \(\left\{\begin{matrix}\frac{1}{x}+y=a\\\frac{y}{x}=b\end{matrix}\right.\) ta dc:

\(\left\{\begin{matrix}ab=6\\a^2-2b=5\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}a=3\\b=2\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\frac{1}{x}+y=3\\\frac{y}{x}=2\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x=\frac{1}{2}\\y=1\end{matrix}\right.\) hoặc \(\left\{\begin{matrix}x=1\\y=2\end{matrix}\right.\)

đúng mà