\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-6xy+2y^2=6\\x^2+2xy-2y^2=6\end{matrix}\right.\)
\(\Rightarrow x^2-8xy+4y^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\left(4+2\sqrt{3}\right)y\\x=\left(4-2\sqrt{3}\right)y\end{matrix}\right.\)
Thế lên trên là được, nhưng chắc bạn ghi nhầm đề nên pt xấu quá
Giải hpt:
a)\(\left\{{}\begin{matrix}x^4+4x^2y+y^2=6x^2\\x^2+x+y=3xy\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2\left(y^2+2\right)+2y\left(x^2+x+1\right)=3\\\left(x^2+x\right)\left(y^2+y\right)=1\end{matrix}\right.\)
giải hệ phương trình
1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)
3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
giải hpt\(\left\{{}\begin{matrix}3\left|x-2y\right|+x^2+2y=6\\\left|x-2y\right|+2\left(x^2+2y\right)=7\end{matrix}\right.\)
Giải hệ
\(\left\{{}\begin{matrix}x^3+y^3=2\\x^2y+3xy^2+2y^3=6\end{matrix}\right.\)
Giải hpt : \(\left\{{}\begin{matrix}x^2+y^2-2y-6+2\sqrt{2y+3}=0\\\left(x-y\right)\left(x^2+xy+y^2+3\right)=3\left(x^2+y^2\right)+2\end{matrix}\right.\)
Giải HPT: \(\left\{{}\begin{matrix}x^2+2y^2=3\\x+y^2+xy=1\end{matrix}\right.\)
1.Giải hpt : a,\(\left\{{}\begin{matrix}\left(x+y+3\right)\sqrt{x-2y}+2y+4=0\\\left(x-y\right)\left(x^2+4\right)=y^2+1\end{matrix}\right.\)
1/ Giải hpt = p đặt ẩn phụ : a,\(\left\{{}\begin{matrix}\left(x+y\right)^3+y=5\\3\left(x+y\right)^3-22xy+21=11x^2+12y^3\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}81x^3y^2-81x^2y^2+33xy^2-29y^2=4\\25y^3+9x^2y^3-6xy^3-4y^2=24\end{matrix}\right.\)
giải hệ phương trình:
1, \(\left\{{}\begin{matrix}\sqrt{3+2x^2y-x^4y^2}+x^2\left(1-2x^2\right)=y^4\\1+\sqrt{1+\left(x-y\right)^2}=-x^2\left(x^4+1-2x^2-2xy^2\right)\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\sqrt{x-1}+\sqrt{x}\left(3\sqrt{x}-y\right)+x\sqrt{x}=3y+\sqrt{y-1}\\3xy^2+4=4x^2+2y+x\end{matrix}\right.\)