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

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

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

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

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

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

Lời giải:

HPT \(\Leftrightarrow \left\{\begin{matrix} x-2=2y^3-6y-4\\ y-2=-x^3+3x+2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-2=(y-2)(y+1)^2\\ y-2=-(x-2)(x+1)^2\end{matrix}\right.\)

$\Rightarrow y-2=-(y-2)(y+1)^2(x+1)^2$

$\Leftrightarrow (y-2)[(y+1)^2(x+1)^2+1]=0$

Dễ thấy biểu thức trong ngoặc vuông luôn lớn hơn $0$. Do đó $y-2=0$

$\Rightarrow y=2$

Thay vào: $x-2=(y-2)(y+1)^2=0\Rightarrow x=2$

Vậy HPT có nghiệm $(x,y)=(2,2)$

@tth_new

ai đó giải giúp bài này đi

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

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

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

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:

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

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

Trừ 2 vế của phương trình, ta được: 

\(x^3-y^3=-5x+5y\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+5\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+5\right)=0\)

\(\Rightarrow x=y\)

Thay vào hệ ban đầu, ta được: \(x^3=3x+8x\)

\(\Leftrightarrow x^3-11x=0\)

\(\Leftrightarrow x\left(x^2-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y=0\\x=y=\pm\sqrt{11}\end{matrix}\right.\)

Vậy...