Ta có : \(\left\{{}\begin{matrix}2x^2+2y^2-xy=5\\2x-2y-xy=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+2y^2-4xy=2\left(x-y\right)^2=5-3xy\\x-y=\dfrac{xy+5}{2}\end{matrix}\right.\)
\(\Rightarrow2\left(\dfrac{xy+5}{2}\right)^2=5-3xy\)
\(\Rightarrow\left(xy\right)^2+10xy+25=10-6xy\)
\(\Leftrightarrow\left(xy\right)^2+16xy+15=0\)
\(\Leftrightarrow\left[{}\begin{matrix}xy=-1\\xy=-15\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-y=2\\x-y=-5\end{matrix}\right.\)
- Đến đây giải nốt bằng phương pháp thế nha bạn
Ta được : \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Vậy ...
a.
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-y\right)^2+3xy=5\\2\left(x-y\right)-xy=5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x-y=u\\xy=v\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}2u^2+3v=5\\2u-v=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2u^2+3v=5\\v=2u-5\end{matrix}\right.\)
\(\Rightarrow2u^2+6u-20=0\Rightarrow\left[{}\begin{matrix}u=2;v=-1\\u=-5;v=-15\end{matrix}\right.\)
-TH1: \(\left\{{}\begin{matrix}u=2\\v=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=2\\xy=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x-2\\xy+1=0\end{matrix}\right.\)
\(\Rightarrow x\left(x-2\right)+1=0\Leftrightarrow x=1\Rightarrow y=-1\)
TH2: \(\left\{{}\begin{matrix}u=-5\\v=-15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-y=-5\\xy+15=0\end{matrix}\right.\)
\(\Leftrightarrow x\left(x+5\right)+15=0\) (vô nghiệm)
b, Ta có : \(\left\{{}\begin{matrix}x+y+xy=3\\x^2+y^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=3-xy\\x^2+2xy+y^2=\left(x+y\right)^2=2+2xy\end{matrix}\right.\)
\(\Rightarrow\left(xy-3\right)^2=2xy+2\)
\(\Leftrightarrow\left(xy\right)^2-6xy+9-2xy-2=0\)
\(\Leftrightarrow\left(xy\right)^2-8xy+7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}xy=7\\xy=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x+y=-4\\x+y=2\end{matrix}\right.\)
- Giair bằng x y là nghiệm của phương trình \(X^2-SX+P=0\)
\(\Rightarrow x=y=1\)
Vậy ...
b.
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=3\\\left(x+y\right)^2-2xy=2\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\) ta được:
\(\left\{{}\begin{matrix}u+v=3\\u^2-2v=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}v=3-u\\u^2-2v=2\end{matrix}\right.\)
\(\Rightarrow u^2-2\left(3-u\right)=2\)
\(\Leftrightarrow u^2+2u-8=0\Rightarrow\left[{}\begin{matrix}u=2\Rightarrow v=1\\u=-4\Rightarrow v=7\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Rightarrow x\left(2-x\right)-1=0\)
\(\Leftrightarrow x=1\Rightarrow y=1\)
Câu b còn 1 cách giải khác là sử dụng BĐT:
Ta có:
\(\left(x^2+1\right)+\left(y^2+1\right)+\left(x^2+y^2\right)\ge2x+2y+2xy\)
\(\Leftrightarrow2\left(x^2+y^2\right)+2\ge2\left(x+y+xy\right)=6\)
\(\Leftrightarrow x^2+y^2\ge2\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=1\)
Vậy hệ có nghiệm duy nhất \(\left(x;y\right)=\left(1;1\right)\)