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Lấy \(pt\left(1\right)-3.pt\left(2\right)\)được
\(11y^2+11y=22\)
\(\Leftrightarrow y^2+y-2=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=-2\end{cases}}\)
Thế vô 1 trong 2 pt đầu sẽ tìm đc x
\(\left\{{}\begin{matrix}5\left(x+2y\right)=4x-1\\2x+4=3\left(x-5y\right)-20\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x+10y=4x-1\\2x+4=3x-15y-20\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+10y=-1\left(1\right)\\x-15y=24\left(2\right)\end{matrix}\right.\)
Lấy (1)-(2): \(25y=-25\Leftrightarrow y=-1\) thay vào (1) \(\Leftrightarrow x=9\)
a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=10\\xy\left(x+y\right)+5\left(x+y\right)=32\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u^2-2v=10\\uv+5u=32\end{matrix}\right.\)
\(\Rightarrow u\left(\dfrac{u^2-10}{2}\right)+5u=32\)
\(\Leftrightarrow u^3=64\Rightarrow u=4\Rightarrow v=3\)
\(\Rightarrow\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)
a)\(\left\{{}\begin{matrix}8x+2y=4\\8x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\4x+1=2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=1\\x=\frac{1}{4}\end{matrix}\right.\)b)
\(\left\{{}\begin{matrix}12x-8y=44\\12x-15y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=35\\4x-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\4x-5.5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)c)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)