b,\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\4\sqrt{x+3}+2\sqrt{y+1}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+3}=10\\\sqrt{x+3}=2\\\sqrt{y+1}=4-2.2\\\sqrt{y+1}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3=4\\x=1\\y=0\left(ktm\right)\\y=-1\end{matrix}\right.\)
Vậy : \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
a,
đkxđ: \(x\ge0;y\ge1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+2\sqrt{y-1}=5\\8\sqrt{x}-2\sqrt{y-1}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}=9\\\sqrt{x}+2\sqrt{y-1}=5\end{matrix}\right.\)
\(\Leftrightarrow x=1;\sqrt{y-1}=2\)
\=> x=1 ; y=5 (tm)
\(a,Đặt:u=\sqrt{x};b=\sqrt{y-1}\left(ĐK:x\ge0;y\ge1\right)\\\Leftrightarrow \left\{{}\begin{matrix}u+2b=5\\4u-b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4u+8b=20\\4a-b=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}9b=18\\4u-b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=2\\4u-2=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=2\left(TM\right)\\u=1\left(TM\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=u^2=1\\\sqrt{y-1}=b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
