Biến đổi phương trình (1) thế vào phương trình (2)
Biến đổi phương trình (1) thế vào phương trình (2)
Giải HPT
1)\(\left\{{}\begin{matrix}x^2+y^2+z=1\\x^2+y+z^2=1\\x+y^2+z^2=1\end{matrix}\right.\)
2)
\(\left\{{}\begin{matrix}xyz=x+y+z\\yzt=y+z+t\\ztx=z+t+x\\txy=t+x+y\end{matrix}\right.\)
3)
\(\left\{{}\begin{matrix}x^3+y^2=2\\x^2+xy+y^2-y=0\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}x^2y^2-2x+y^2=0\\2x^2-4x+y^3+3=0\end{matrix}\right.\)
Giải các hệ phương trình
a) \(\left\{{}\begin{matrix}x+y+xy=3\\x^2y+xy^2=2\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}x^2+y^2=2\left(xy+2\right)\\x+y=6\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x^2-2x=y\\y^2-2y=x\end{matrix}\right.\) d) \(\left\{{}\begin{matrix}2x^2-xy+3y^2=13\\x^2+4xy-2t^2=-6\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}2x^2-y^2=1\\xy+x^2=2\end{matrix}\right.\)
Giải các hệ phương trình
a) \(\left\{{}\begin{matrix}x+y+xy=3\\x^2y+xy^2=2\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}x^2+y^2=2\left(xy+2\right)\\x+y=6\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x^2-2x=y\\y^2-2y=x\end{matrix}\right.\) d) \(\left\{{}\begin{matrix}2x^2-xy+3y^2=13\\x^2+4xy-2y^2=-6\end{matrix}\right.\)
e) \(\left\{{}\begin{matrix}2x^2-y^2=1\\xy+x^2=2\end{matrix}\right.\) f) \(\left\{{}\begin{matrix}x^2-y^2=1-xy\\x^2+y^2=3xy+11\end{matrix}\right.\)
Cần gấp lắm, ai giúp với
Giải hệ pt:
a)\(\left\{{}\begin{matrix}x^2+y^2+x+y=18\\x\left(x+1\right).y\left(y+1\right)=72\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\3y-1=xy\end{matrix}\right.\) c)\(\left\{{}\begin{matrix}2x+3y=xy+5\\\frac{1}{x}+\frac{1}{y+1}=1\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\sqrt{\frac{x}{y}}-3\sqrt{\frac{y}{x}}=2\\x-y+xy=1\end{matrix}\right.\) e)\(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
HELP ME :((
Giải HPT: \(\left[{}\begin{matrix}\sqrt{x}\left(1+y\right)=2y\\\sqrt{y}\left(1+z\right)=2z\\\sqrt{z}\left(1+x\right)2x\end{matrix}\right.\)
Giải hệ phương trình :
a, \(\left\{{}\begin{matrix}\left(x+y-2\right)\left(2x-y\right)=0\\x^2+y^2=2\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}x^2+y^2+2x+2y=6\\x+y-3xy+1=0\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}x^2+4x=5y\\y^2+4y=5x\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}x^2+2y^2+xy=4\\2x^2+xy+3y^2=6\end{matrix}\right.\)
e,\(\left\{{}\begin{matrix}4x^2+8x=5y\\y^2+4y=10x\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2x^2}{1+x^2}=y\\\dfrac{2y^2}{1+y^2}=z\\\dfrac{2z^2}{1+z^2}=x\end{matrix}\right.\)
Giải các hệ phương trình sau
a)\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\2x+3y=xy+5\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\left(x-y\right)^2+3\left(x-y\right)=4\\2x+3y=12\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\frac{x}{y}+\frac{y}{x}=\frac{13}{6}\\x+y=5\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}x+y+xy=7\\x+y^2+xy=13\end{matrix}\right.\)
Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}x^2+xy+y^2=4\\x+xy+y=2\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}x^3=2y+1\\y^3=2x+1\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x+y-\sqrt{xy}=3\\\sqrt{x+1}+\sqrt{y+1}=4\end{matrix}\right.\)