Giai hệ PT sau:\(\left\{{}\begin{matrix}2x^2+xy=3y+6\\2y^2+xy=3x+6\end{matrix}\right.\)
\(\left\{{}\begin{matrix}xy+x^2=1+y\\yx+y^2=1+x\end{matrix}\right.\)
hệ phương trình
1, \(\left\{{}\begin{matrix}3x=6\\x-3y=2\end{matrix}\right.\)
2,\(\left\{{}\begin{matrix}3x+5y=15\\2y=-7\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}7x-2y=1\\3x+y=6\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}3\left(x+y\right)+9=2\left(x-y\right)\\2\left(x+y\right)=3\left(x-y\right)+11\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}3\left(x+y\right)+5\left(x-y\right)=12\\-5\left(x+y\right)+2\left(x-y\right)=11\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}2\left(3x-2\right)-4=5\left(3y+2\right)\\4\left(3x-2\right)+7\left(3y+2\right)=-2\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=\frac{4}{5}\\\frac{1}{x}-\frac{1}{y}=\frac{1}{5}\end{matrix}\right.\)
8 , \(\left\{{}\begin{matrix}\frac{15}{x}-\frac{7}{y}=9\\\frac{4}{x}+\frac{9}{y}=35\end{matrix}\right.\)
Giải pt sau bằng cách đặt ẩn phụ
1, \(\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=-\frac{5}{4}\\x^4+y^2+xy\left(1+2x\right)=-\frac{5}{4}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^3+3x^2-13x-15=\frac{8}{y^3}-\frac{8}{y}\\y^2+4=5y^2\left(x^2+2x+2\right)\end{matrix}\right.\)
Giải các hệ
\(\left\{{}\begin{matrix}\sqrt{x+y}+\sqrt{2x+y+2}=7\\3x+2y=23\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=\frac{-5}{4}\\x^4+y^2+xy\left(1+2x\right)=\frac{-5}{4}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x^2+1\right)+y\left(x+y\right)=7y\\\left(x^2+1\right)\left(x+y-2\right)=-y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x\left(x+y+1\right)=3\\\left(x+y\right)^2-\frac{5}{x^2}=-1\end{matrix}\right.\)
giải hệ Pt : \(\left\{{}\begin{matrix}x^2\left(y+1\right)\left(x+y+1\right)=3x^2-4x+1\\xy+x+1=x^2\end{matrix}\right.\)
Tìm giá trị của m để các hệ phương trình sau vô nghiệm ?
a) \(\left\{{}\begin{matrix}3x+2y=9\\mx-2y=2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x-my=5\\x+y=7\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2x^2-y^2+xy-3x+3y+2=0\\x^2+y^2=2\end{matrix}\right.\)
Giải các hệ phương trình :
a. \(\left\{{}\begin{matrix}x+3y+2z=8\\2x+2y+z=6\\3x+y+z=6\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}x-3y+2z=-7\\-2x+4y+3z=8\\3x+y-z=5\end{matrix}\right.\)
