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

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

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

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?

a) x(x-1)=0+12

    x(x-1)=12

    x(x-1)=4.3

=>x=4

a, \(x^2-x-12=0\)

\(x^2+\left(-x\right)+\left(-12\right)=0\)

\(\Delta=-1^2-4.1.\left(-12\right)=1+48=49>0\)

Nên pt có 2 nghiệm phân biệt 

\(x_1=\frac{1-\sqrt{49}}{2.1}=\frac{1-7}{2}=-\frac{6}{2}=-3\)

\(x_2=\frac{1+\sqrt{49}}{2.1}=\frac{1+7}{2}=\frac{8}{2}=4\)

ĐKXĐ: ...

\(x\sqrt{x}-3x\sqrt{y}+3y\sqrt{x}-y\sqrt{y}=0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^3=0\)

\(\Leftrightarrow\sqrt{x}=\sqrt{y}\Leftrightarrow x=y\)

Thay xuống dưới:

\(x^2-2x+x=0\Leftrightarrow x^2-x=0\)

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

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...

\(\sqrt{2x+1}-\sqrt{3x}=x-1\)

ĐK: \(x\ge0\)

\(\sqrt{2x+1}-\sqrt{3x}=3x-\left(2x+1\right)\)

\(\Leftrightarrow\sqrt{2x+1}-\sqrt{3x}=\left(\sqrt{3x}-\sqrt{2x+1}\right)\left(\sqrt{3x}+\sqrt{2x+1}\right)\)

\(\Leftrightarrow\left(\sqrt{2x+1}-\sqrt{3x}\right)\left(1+\sqrt{3x}+\sqrt{2x+1}\right)=0\)

\(\Leftrightarrow\sqrt{2x+1}=\sqrt{3x}\Rightarrow x=1\left(tm\right)\)

ai giải hộ mk ý a vs ý c

c) \(x^3y+xy^3-3x^2-3y^2=17\)

\(\Leftrightarrow xy\left(x^2+y^2\right)-3\left(x^2+y^2\right)=17\Leftrightarrow\left(x^2+y^2\right)\left(xy-3\right)=17\)

\(\Leftrightarrow\left(x^2+y^2\right),\left(xy-3\right)\inƯ\left(17\right)\)

Do \(x^2+y^2\ge0\Rightarrow x^2+y^2\in\left\{1;17\right\}\)

TH1:  \(\hept{\begin{cases}x^2+y^2=1\\xy-3=17\end{cases}}\Rightarrow\hept{\begin{cases}\frac{400}{y^2}+y^2=1\\x=\frac{20}{y}\end{cases}}\) (vô nghiệm)

TH2:  \(\hept{\begin{cases}x^2+y^2=17\\xy-3=1\end{cases}}\Rightarrow\hept{\begin{cases}\frac{16}{y^2}+y^2=17\\x=\frac{4}{y}\end{cases}}\)

Ta có bảng:

Vậy các cặp số nguyên thỏa mãn là (x;y) = (1;4) ; (-1;-4) ; (4;1) ; (-4;-1).

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

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

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

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

Thay xuống pt dưới ...