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

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

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

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?

ý a ở đây bn https://hoc247.net/hoi-dap/toan-10/giai-he-pt-3x-x-2-2-y-2-va-3y-y-2-2-x-2-faq371128.html

b.

Với \(xy=0\) không là nghiệm

Với \(xy\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y^2+1\right)=y\left(5-x^2\right)\\y^2+1=y\left(5-2x\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y^2+1}{y}=\dfrac{5-x^2}{x}\\\dfrac{y^2+1}{y}=5-2x\end{matrix}\right.\)

\(\Rightarrow\dfrac{5-x^2}{x}=5-2x\)

\(\Leftrightarrow5-x^2=5x-2x^2\)

\(\Leftrightarrow...\)

c.

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

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

\(\Rightarrow5x^2-x\left(y+1\right)-4\left(y+1\right)^2=0\)

\(\Leftrightarrow\left(x-y-1\right)\left(5x+4\left(y+1\right)\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=x-1\\y=-\dfrac{5x+4}{4}\end{matrix}\right.\)

Thế vào 1 trong 2 pt ban đầu...

Với \(xy=0\) là nghiệm

Với \(xy\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}y-\dfrac{2}{x}+\dfrac{3x}{y}=0\\\dfrac{y}{x}+x+\dfrac{2}{y}=0\end{matrix}\right.\)

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

\(\Rightarrow\left(y-\dfrac{2}{x}\right)\left(x+\dfrac{2}{y}\right)=3\)

\(\Leftrightarrow xy-\dfrac{4}{xy}-3=0\)

\(\Rightarrow\left(xy\right)^2-3xy-4=0\Rightarrow\left[{}\begin{matrix}xy=-1\\xy=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{y}\\x=\dfrac{4}{y}\end{matrix}\right.\) thế vào \(y^2+x^2y+2x=0\)

\(\Rightarrow\left[{}\begin{matrix}y^2+\dfrac{1}{y}-\dfrac{2}{y}=0\\y^2+\dfrac{16}{y}+\dfrac{8}{y}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y^3=1\\y^3=-24\end{matrix}\right.\)

\(\Leftrightarrow...\)

nhìn mà chóng mặt nhưng để mk thử xem 

ngu

\(\int^{\text{2x2+x−1y=2y−y2x−2y2=−2}}_{\text{2x2+x−1y=2y−y2x−2y2=−2}}\)

ĐKXĐ: y≠0

\(\int^{2x^2+x-\frac{1}{y}\left(1\right)=2}_{y-y^2x-2y^2=-2}\)

Do y≠0 nên chia phương trình 2 cho y²

<=>\(\int^{\text{2x2+x−1y=2}}_{1y−x−2=−2y2}\)<=>\(\int^{2x^2+x-\frac{1}{y}=2}_{2+x-\frac{1}{y}=\frac{2}{y^2}}\)

Trừ 2 phương trình, rút gọn, ta được:

x²−1=1−\(\frac{1}{y^2}\)

<=>\(\frac{1}{y}=\sqrt{2-x^2}\)

Thay vào (1), ta được phương trình sau:
\(2x^2+x-\sqrt{2-x^2}=2\)

<=>\(-2\left(2-x^2\right)+x+2=\sqrt{2-x^2}\left(2\right)\)

Đặt \(a=\sqrt{2-x^2}\)

=>a²=2−x²

=>a²+x²=2

Thay vào (2), ta được phương trình:

2a²+x+a²+x²=a

<=>x²−a²+x−a=0

<=>(x−a)(x+a+1)=0

<=>\(\int^{x-a=0}_{x+a+1=0}\)

TH1: x−a=0=>x=a

=>\(x=\sqrt{2-x^2}\)

=>x=±1

Với x=1, thay vào (1), tính được y=1

Với x=−1, thay vào (1), tính được y=−1

TH2:x+a+1=0=>x+1=-a

=>\(x+1=-\sqrt{2-x^2}\)

ĐK: x≤\(-\sqrt{2}\)

\(x+1=-\sqrt{2-x^2}\)

<=>x²+2x+1=2-x²

<=>2x²+2x-1=0

<=>\(\int^{\frac{\sqrt{3}-1}{2}\left(loại\right)}_{\frac{-\sqrt{3}-1}{2}\left(loại\right)}\)


KẾT LUẬN: Hệ phương trình có 2 cặp nghiệm (x;y):(1;1),(−1;−1)

Cộng vế:

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

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

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

\(\Leftrightarrow\left(x-y+2\right)\left[\left(x+\dfrac{y}{2}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2+1\right]=0\)

\(\Leftrightarrow y=x+2\)

1.

HPT  \(\left\{\begin{matrix} (x+1)(y-1)=xy+4\\ (2x-4)(y+1)=2xy+5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} xy-x+y-1=xy+4\\ 2xy+2x-4y-4=2xy+5\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} -x+y=5\\ 2x-4y=9\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x=\frac{-29}{2}\\ y=\frac{-19}{2}\end{matrix}\right.\)

Vậy.............

2.

ĐKXĐ: $x\in\mathbb{R}$

$x^2+x-2\sqrt{x^2+x+1}+2=0$

$\Leftrightarrow (x^2+x+1)-2\sqrt{x^2+x+1}+1=0$

$\Leftrightarrow (\sqrt{x^2+x+1}-1)^2=0$

$\Rightarrow \sqrt{x^2+x+1}=1$

$\Rightarrow x^2+x=0$

$\Leftrightarrow x(x+1)=0$

$\Rightarrow x=0$ hoặc $x=-1$