1) \(\left\{{}\begin{matrix}3\sqrt{x-2}+\sqrt{y}=5\\5\sqrt{x-2}-\sqrt{y}=3\end{matrix}\right.\left(x\ge2;y\ge0\right)\)
Đặt: \(\left\{{}\begin{matrix}\sqrt{x-2}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a+b=5\\5a-b=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8a=8\\5a-b=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\5\cdot1-b=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\left(tm\right)\)
1: ĐKXĐ: x>=2 và y>=0
\(\left\{{}\begin{matrix}3\sqrt{x-2}+\sqrt{y}=5\\5\sqrt{x-2}-\sqrt{y}=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3\sqrt{x-2}+\sqrt{y}+5\sqrt{x-2}-\sqrt{y}=5+3=8\\5\sqrt{x-2}-\sqrt{y}=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8\sqrt{x-2}=8\\5\sqrt{x-2}-\sqrt{y}=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y}=5-3=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2=1\\y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\left(nhận\right)\)
2: Để hệ có nghiệm duy nhất thì \(\dfrac{2}{m}\ne\dfrac{1}{2}\)
=>\(m\ne4\)
\(\left\{{}\begin{matrix}2x+y=1\\mx+2y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+2y=2\\mx+2y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}mx+2y-4x-2y=3-2\\2x+y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(m-4\right)=1\\y=1-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{m-4}\\y=1-\dfrac{2}{m-4}=\dfrac{m-6}{m-4}\end{matrix}\right.\)
Để \(P=3x+y=\dfrac{3}{m-4}+\dfrac{m-6}{m-4}=\dfrac{m-3}{m-4}\in Z\) thì \(m-3⋮m-4\)
=>\(m-4+1⋮m-4\)
=>\(1⋮m-4\)
=>\(m-4\in\left\{1;-1\right\}\)
=>\(m\in\left\{3;5\right\}\)
