Xét hệ phương trình: \(\left\{{}\begin{matrix}y+xy^2=6x^2\left(1\right)\\1+x^2y^2=5x^2\left(2\right)\end{matrix}\right.\)

Từ (2) => x # 0

Chia 2 vế của mỗi PT cho x² ta được \(\left\{{}\begin{matrix}\dfrac{y}{x^2}+\dfrac{y^2}{x}=6\\\dfrac{1}{x^2}+y^2=5\end{matrix}\right.\)

Đặt \(a=\dfrac{1}{x}\) ta có \(\left\{{}\begin{matrix}a^2y+ay^2=6\\a^2+y^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}ay\left(a+y\right)=6\\\left(a+y\right)^2-2ay=5\end{matrix}\right.\)

Đặt t = a + y, z =ay (t² \(\ge\) 4z)

Ta có: \(\left\{{}\begin{matrix}tz=6\\t^2-2z=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=\dfrac{t^2-5}{2}\\t^3-5t-12=0\left(3\right)\end{matrix}\right.\)

(3) <=> (t - 3)(t² + 3t + 4) = 0 <=> t = 3 => z = 2

Vậy \(\left\{{}\begin{matrix}a+y=3\\a.y=2\end{matrix}\right.\)

\(\Leftrightarrow\left(a=1;y=2\right)\) hoặc \(\left(a=2;y=1\right)\)

Hệ thức có hai nghiệm (x = 1; y = 2), (x = \(\dfrac{1}{2}\) ; x = 1)

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

Tham khảo nha:

https://hoc247.net/hoi-dap/toan-9/giai-he-phuong-trinh-y-xy-2-6x-2-1-x2y-2-5x-2-faq361806.html

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

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

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

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?

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=10\\xy\left(x+y\right)+5\left(x+y\right)=32\end{matrix}\right.\)

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

\(\Rightarrow\left\{{}\begin{matrix}u^2-2v=10\\uv+5u=32\end{matrix}\right.\)

\(\Rightarrow u\left(\dfrac{u^2-10}{2}\right)+5u=32\)

\(\Leftrightarrow u^3=64\Rightarrow u=4\Rightarrow v=3\)

\(\Rightarrow\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)

Lời giải:

Nếu \(x=0\Rightarrow x^2y^2=-1\) (vô lý)

Nếu \(y=0\Rightarrow 6x^2=0\Leftrightarrow x=0\).Thay vào pt (2) thì \(1=5x^2=0\) (vô lý)

Vậy \(x,y\neq 0\)

PT tương đương: \(\left\{\begin{matrix} y(1+xy)=6x^2\\ (xy+1)^2-2xy=5x^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} xy+1=\frac{6x^2}{y}\\ (xy+1)^2-2xy=5x^2\end{matrix}\right.\)

\(\Rightarrow \left(\frac{6x^2}{y}\right)^2-2xy=5x^2\)

\(\Leftrightarrow \frac{36x^3}{y^2}-2y=5x\) (do \(x\neq 0\) )

\(\Leftrightarrow 36x^3-2y^3=5xy^2\)

Đặt \(x=ty\Rightarrow 36t^3y^3-2y^3-5ty^3=0\)

\(\Leftrightarrow 36t^3-2-5t=0\) (do \(y\neq 0\) )

\(\Leftrightarrow (2t-1)(18t^2+9t+2)=0\)

Thấy rằng \(18t^2+9t+2=18(t+\frac{1}{4})^2+\frac{7}{8}>0\) nên \(2t-1=0\)

\(\Leftrightarrow t=\frac{1}{2}\Leftrightarrow x=\frac{y}{2}\Leftrightarrow 2x=y\)

Thay vào PT (1)

\(2x+4x^3=6x^2\Leftrightarrow 1+2x^2-3x=0\) (do x khác 0)

\(\Leftrightarrow (2x-1)(x-1)=0\)

Nếu \(x=\frac{1}{2}\Rightarrow y=1\)

Nếu \(x=1\Rightarrow y=2\)

Thử lại thấy thỏa mãn.

Vậy \((x,y)\in \left\{(\frac{1}{2};1); (1;2)\right\}\)