Các câu hỏi tương tự
\(\left\{{}\begin{matrix}x^2+y^2+2x+4y=8\\\left(x+2y+1\right)\left(9+3y^2+4xy\right)=64\end{matrix}\right.\)
26 tháng 10 2021 lúc 17:19
Ghpt:
a) \(\left\{{}\begin{matrix}x^2+2y^2=2x-2xy+1\\3x^2+2xy-y^2=2x-y+5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}4xy+4x^2+4y^2+\dfrac{3}{\left(x+y\right)^2}=7\\2x+\dfrac{1}{x+y}=3\end{matrix}\right.\)
Giải các hệ phương trình:a)left{{}begin{matrix}dfrac{x}{y}dfrac{2}{3}x+y-100end{matrix}right.b)left{{}begin{matrix}left(3x+2right)left(2y-3right)6xyleft(4x+5right)left(y-5right)4xyend{matrix}right.c)left{{}begin{matrix}left(2x-3right)left(2y+4right)4xleft(y-3right)+54left(x+1right)left(3y-3right)3yleft(x+1right)-12end{matrix}right.d)left{{}begin{matrix}dfrac{2y-5x}{3}+5dfrac{y+27}{4}-2xdfrac{x+1}{3}+ydfrac{6y-5x}{7}end{matrix}right.
Đọc tiếp
Giải các hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\left(3x+2\right)\left(2y-3\right)=6xy\\\left(4x+5\right)\left(y-5\right)=4xy\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{2y-5x}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+3y^2-4xy-x+3y=0\\4xy+3x+2y=-2\end{matrix}\right.\)giải hệ pt sau
\(\left\{{}\begin{matrix}x^2-5xy-2y^2=3\\x^2+4xy+4=0\end{matrix}\right.\)
30 tháng 11 2021 lúc 21:09
Giải hệ \(\left\{{}\begin{matrix}\sqrt{x^2+2y+3}+2y-3=0\\2\left(2y^3+x^3\right)+3y\left(x+1\right)^2+6\left(x+1\right)+2=0\end{matrix}\right.\)
2 tháng 12 2021 lúc 20:38
Giải hệ\(\left\{{}\begin{matrix}\sqrt{x^2+2y+3}+2y-3=0\\2\left(2y^3+x^3\right)+3y\left(x+1\right)^2+6\left(x+1\right)+2=0\end{matrix}\right.\)
1 tháng 12 2021 lúc 15:51
Giải hệ:\(\left\{{}\begin{matrix}\sqrt{x^2+2y+3}+2y-3=0\\2\left(2y^3+x^3\right)+3y\left(x+1\right)^2+6\left(x+1\right)+2=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+2y^2-3=0\\x\left(x^2+3\right)-4y^3=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}4xy+x+4\sqrt{\left(2-x\right)\left(2+y\right)}=14\\x^2+y^2+2x-1=0\end{matrix}\right.\)