Cộng 2 pt lại ta được

\(x^2+y^2+2xy-4x-4y=-3\)

\(\Leftrightarrow\left(x+y\right)^2-4\left(x+y\right)+3=0\)

\(\Leftrightarrow\left(x+y-1\right)\left(x+y-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+y=1\\x+y=3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1-y\\x=3-y\end{cases}}\)

THế vào 1 trong 2 pt ban đầu là Ok

\(\left\{{}\begin{matrix}\left(x+2y\right)^2=5+4xy\\\left(x+2y\right)\left(5+4xy\right)=27\end{matrix}\right.\)

\(\Rightarrow\left(x+2y\right)^3=27\Rightarrow x+2y=3\Rightarrow x=3-2y\)

Thay vào pt đầu:

\(\left(3-2y\right)^2+4y^2-5=0\)

\(\Leftrightarrow8y^2-12y+4=0\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=1\\y=\frac{1}{2}\Rightarrow x=2\end{matrix}\right.\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}x^3-2xy^2+y\left(x^2-8y^2\right)=0\\x^2-8y^2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2y\right)\left(x^2+xy+4y^2\right)=0\\x^2-8y^2=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2y\\x^2+xy+4y^2=0\end{matrix}\right.\\x^2-8y^2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y\\x^2-8y^2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y\\\left(2y\right)^2-8y^2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2;y=1\\x=-2;y=-1\end{matrix}\right.\).

Phương trình (1) có hai cái x^2 là sao?

\(pt(2)<=>(2x+y-8)(2x+y+7)=0\)

pt <=> \(\hept{\begin{cases}x^2-4y^2-8x+4y+15=0\\3x^2+6y^2-6xy=15\end{cases}}\)

\(\hept{\begin{cases}x^2+2y^2-2xy=5\\4x^2+2y^2-6xy-8x+4y=0\end{cases}}\)

\(\hept{\begin{cases}x^2+2y^2-2xy=5\\\left(2x-y\right)\left(x-y-2\right)=0\end{cases}}\)

tới đây bạn giải quyết được rồi nhé

nhanh đy mằ =(

ĐKXĐ : \(\left\{{}\begin{matrix}x\ge2\\y\ge-\dfrac{1}{2}\end{matrix}\right.\)

Ta có \(\left\{{}\begin{matrix}x+2y-1-2\sqrt{2xy+x-4y-2}=0\\\sqrt{x-2}+3\sqrt{2y+1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)+\left(2y+1\right)-2\sqrt{\left(x-2\right)\left(2y+1\right)}=0\\\sqrt{x-2}+3\sqrt{2y+1}=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x-2}-\sqrt{2y+1}\right)^2=0\\\sqrt{x-2}+3\sqrt{2y+1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{2y+1}\\4\sqrt{2y+1}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{2y+1}\\2y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{2y+1}\\y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=0\end{matrix}\right.\)