Cộng vế với vế:

\(4x^2-4xy+y^2=0\Leftrightarrow\left(2x-y\right)^2=0\Leftrightarrow2x=y\)

Thay vào pt đầu:

\(4x^2-x.2x=2\) \(\Leftrightarrow2x^2=2\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=2\\x=-1\Rightarrow y=-2\end{matrix}\right.\)

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

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

Đặt \(\left\{{}\begin{matrix}x^2-2x=u\\y^2-4y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2u-v=1\\u^2+2=uv\end{matrix}\right.\) \(\Rightarrow u^2+2=u\left(2u-1\right)\)

\(\Leftrightarrow u^2-u-2=0\Leftrightarrow...\)

Cộng vế với vế:

\(x^2+2xy+y^2+x+y=12\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+y=-4\\x+y=3\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=-4\\xy=5-\left(x+y\right)=9\end{matrix}\right.\)

Theo Viet đảo, x và y là nghiệm: \(t^2-4t+9=0\) (vô nghiệm)

TH2: \(\left\{{}\begin{matrix}x+y=3\\xy=5-\left(x+y\right)=2\end{matrix}\right.\)

Theo Viet đảo, x và y là nghiệm:

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

\(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+y^2+y=8\\\left(x^2+x\right)\left(y^2+y\right)=12\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2+x=u\\y^2+y=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u+v=8\\uv=12\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(6;2\right);\left(2;6\right)\)

TH1: \(\left\{{}\begin{matrix}x^2+x=6\\y^2+y=2\end{matrix}\right.\) \(\Rightarrow...\)

TH2: ... tương tự

cảm ơn thầy ạ 3>

Tru (1) cho (2) , ta dc:

x²-y²=4y-4x

⇔(x-y)(x+y)=-4(x-y)

⇔(x-y)(x+y+4)=0

⇔[x=y ; x=-4-y

+) Vs x=y the vao (1)

y²-3y²=4y

⇔[y=0 => x=0 ; y=-2 => x=-2

+) Vs x=-4-y the (2)

y²-3(-4-y)y=4(-4-y)

⇔y=-2 =>x=-2

Xét \(y=0\)\(\Rightarrow...\)

Xét \(y\ne0\). Ta có:

\(\left\{{}\begin{matrix}x^2+y^2+xy+2x=5y\\\left(x^2+2x\right)\left(x+y-3\right)=-3y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x=5y-y^2-xy\left(1\right)\\\left(x^2+2x\right)\left(x+y-3\right)=-3y\left(2\right)\end{matrix}\right.\)

Thay (1) vào (2), ta có:

\(\left(5y-y^2-xy\right)\left(x+y-3\right)=-3y\)

\(-y\left(x+y-5\right)\left(x+y-3\right)=-3y\)

\(\Leftrightarrow\left(x+y-5\right)\left(x+y-3\right)=3\left(\cdot\right)\)

Đặt \(x+y-5=t\), phương trình \(\left(\cdot\right)\) trở thành

\(t\left(t+2\right)=3\)\(\Leftrightarrow t^2+2t+1=4\Leftrightarrow\left(t+1\right)^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}t+1=2\\t+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y-5=1\\x+y-5=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=6\\x+y=2\end{matrix}\right.\)\(\Rightarrow...\)

ĐKXĐ : \(x;y\ne0\)

Ta có \(\dfrac{y}{x}-\dfrac{2x}{y}=\dfrac{-5}{2}-\dfrac{2}{xy}\)

\(\Leftrightarrow\dfrac{y^2-2x^2}{xy}=\dfrac{-5xy-4}{2xy}\)

\(\Leftrightarrow2y^2-4x^2+5xy=-4\) (1) 

Kết hợp \(x^2+xy-y^2=5\) (2)

ta có : \(-5.\left(2y^2-4x^2+5xy\right)=4\left(x^2+xy-y^2\right)\) 

\(\Leftrightarrow16x^2-29xy-6y^2=0\)

\(\Leftrightarrow16x^2-32xy+3xy-6y^2=0\)

\(\Leftrightarrow\left(x-2y\right)\left(16x+3y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2y\\x=-\dfrac{3y}{16}\end{matrix}\right.\)

Thay \(x=-\dfrac{3y}{16}\) vào (2) ta được 

\(\dfrac{9y^2}{256}-\dfrac{3y^2}{16}-y^2=5\)

\(\Leftrightarrow y^2=-\dfrac{256}{59}\Leftrightarrow y\in\varnothing\) (loại) 

Khi x = 2y thay vào (2) ta được 

4y² + 2y² - y² = 5

\(\Leftrightarrow y=\pm1\) (tm)

Với y = 1 => x = 2

y = -1 => x = -2

Vậy (x;y) = (2;1) ; (-2;-1) 

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