\(\Leftrightarrow\left\{{}\begin{matrix}x^3+3xy^2=-49\\3x^2-24xy+3y^2=24y-51x\end{matrix}\right.\)

Cộng vế:

\(x^3+3x^2+3y^2\left(x+1\right)-24y\left(x+1\right)+51x+49=0\)

\(\Leftrightarrow\left(x+1\right)^3+3y^2\left(x+1\right)-24y\left(x+1\right)+48\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^3+3\left(x+1\right)\left(y-4\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)^2+3\left(y-2\right)^2\right]=0\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-2\\y\le\frac{16}{3}\end{matrix}\right.\)

\(2x^2-\left(3y-6\right)x+y^2-8y-20=0\)

\(\Delta=\left(3y-6\right)^2-8\left(y^2-8y-20\right)=y^2+28y+196=\left(y+14\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{3y-6+y+14}{4}=y+2\\x=\frac{3y-6-y-14}{4}=\frac{y-10}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=x-2\\y=2x+10\end{matrix}\right.\)

- Với \(y=2x+10\ge-2.2+10=6>\frac{16}{3}\) ko phù hợp ĐKXĐ (loại)

- Với \(y=x-2\)

\(4\sqrt{x+2}+\sqrt{22-3x}=x^2+8\)

\(\Leftrightarrow x^2+8-4\sqrt{x+2}-\sqrt{22-3x}=0\)

\(\Leftrightarrow x^2-x-2+\frac{4}{3}\left(x+4-3\sqrt{x+2}\right)+\frac{1}{3}\left(14-x-3\sqrt{22-3x}\right)=0\)

\(\Leftrightarrow x^2-x-2+\frac{4}{3}\left(\frac{x^2-x-2}{x+4+3\sqrt{x+2}}\right)+\frac{1}{3}\left(\frac{x^2-x-2}{14-x+3\sqrt{22-3x}}\right)=0\)

\(\Leftrightarrow\left(x^2-x-2\right)\left(....\right)=0\) (ngoặc phía sau luôn dương)

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

a: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\cdot\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=1+2=3\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}3x+2y=2\\6x-3y=18\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=2-2y\\2\cdot3x-3y=18\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=2-2y\\2\left(2-2y\right)-3y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4-7y=18\\3x=2-2y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7y=-14\\3x=2-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\3x=2-2\cdot\left(-2\right)=6\end{matrix}\right.\)

=>x=2 và y=-2

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

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

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

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

Xét (1) \(\Leftrightarrow y=2-x\) thay vào pt đầu: ....

Xét (2): kết hợp với pt đầu ta được:

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

Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\)

\(\Rightarrow\left\{{}\begin{matrix}a+b+2=0\\a^3-3ab-3b=-1\end{matrix}\right.\)

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

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

\(\Leftrightarrow...\)