\(a,\Leftrightarrow\left\{{}\begin{matrix}10x-5-4x+4y=0\\2x-2y-3x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+4y=5\\x+3y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}18x+12y=15\\4x+12y=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14x=27\\x+3y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{27}{14}\\y=-\dfrac{23}{14}\end{matrix}\right.\)
\(f,ĐK:x\ge-4;y\ge1\)
Đặt \(\sqrt{x+4}=a;\sqrt{y-1}=b\left(a,b\ge0\right)\), hpt trở thành:
\(\left\{{}\begin{matrix}2a-b=1\\a+2b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a-b=1\\2a+4b=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a-b=1\\5b=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{6}{5}\\b=\dfrac{7}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+4=\dfrac{36}{25}\\y-1=\dfrac{49}{25}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{64}{25}\left(tm\right)\\y=\dfrac{74}{25}\left(tm\right)\end{matrix}\right.\)
Vậy pt có nghiệm \(\left(x;y\right)=\left(-\dfrac{64}{25};\dfrac{74}{25}\right)\)
