\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\)  \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)

TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)

Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)

TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)

2 câu dưới hình như em hỏi rồi?

b: \(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x-3y-3=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x=3y+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(3y+3\right)^2+y^2-2\left(3y+3\right)-2y-23=0\\x=3y+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9y^2+18y+9+y^2-6y-6-2y-23=0\\x=3y+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}10y^2+10y-20=0\\x=3y+3\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}y\in\left\{-2;1\right\}\\x=3y+3\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\left\{\left(-3;-2\right);\left(6;1\right)\right\}\)

a: \(\left\{{}\begin{matrix}3x^2+6xy-x+3y=0\\4x-9y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9y=4x-6\\3x^2+6xy-x+3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{9}x-\dfrac{2}{3}\\3x^2+6x\cdot\left(\dfrac{4}{9}x-\dfrac{2}{3}\right)-x+3\cdot\left(\dfrac{4}{9}x-\dfrac{2}{3}\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x^2+\dfrac{8}{3}x^2-4x-x+\dfrac{4}{3}x-2=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{17}{3}x^2-\dfrac{11}{3}x-2=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17x^2-11x-6=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x-1\right)\left(17x+6\right)=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}17x+6=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=\dfrac{4}{9}\cdot1-\dfrac{2}{3}=\dfrac{4}{9}-\dfrac{2}{3}=-\dfrac{2}{9}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{6}{17}\\y=\dfrac{4}{9}\cdot\dfrac{-6}{17}-\dfrac{2}{3}=\dfrac{-14}{17}\end{matrix}\right.\end{matrix}\right.\)

Câu 1:

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=3y^2+9\\3x^2+3y^2=3x+12y\end{matrix}\right.\)

\(\Rightarrow x^3-y^3-3x^2-3y^2=3y^2+9-3x-12y\)

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)

\(\Leftrightarrow\left(x-1\right)^3=\left(y+2\right)^3\)

\(\Leftrightarrow x-1=y+2\Rightarrow x=y+3\)

Thay vào pt dưới:

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

\(\Leftrightarrow2y^2+9y+6=0\) \(\Rightarrow...\)

Câu 2:

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2xy+2y^2+3x=0\\2xy+2y^2+6y+2=0\end{matrix}\right.\)

\(\Leftrightarrow x^2+4xy+4y^2+3x+6y+2=0\)

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

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

TH1: \(x+2y=-1\Rightarrow x=-2y-1\) thay vào pt dưới:

\(\left(-2y-1\right)y+y^2+3y+1=0\)

\(\Leftrightarrow-y^2+2y+1=0\Rightarrow...\)

TH2: \(x+2y=-2\Rightarrow x=-2y-2\) thay vào pt dưới:

\(\left(-2y-2\right)y+y^2+3y+1=0\)

\(\Leftrightarrow-y^2-y+1=0\Rightarrow...\)

a/ \(\left\{{}\begin{matrix}x+2y=4\\x^2+4y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\\left(4-2y\right)^2+4y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\4y^2-12y+8=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\\left(y-1\right)\left(y-2\right)=0\end{matrix}\right.\)

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

Vậy hệ phương trình đã cho có nghiệm \(\left(x;y\right)=\left(2;1\right)\) hoặc \(\left(x;y\right)=\left(0;2\right)\)