Violympic toán 9
Các câu hỏi tương tự
12 tháng 12 2020 lúc 7:44
Giải hệ
1) \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2x^2-xy-1=0\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}y\left(4x^3+1\right)=3\\y^3\left(3x-1\right)=4\end{matrix}\right.\)
Giải hệ pt
1/left{{}begin{matrix}4xsqrt{y+1}+8xleft(4x^2-4x-3right)sqrt{x+1}dfrac{x}{x+1}+x^2left(y+2right)sqrt{left(x+1right)left(y+1right)}end{matrix}right.
2/left{{}begin{matrix}xsqrt{y^2+6}+ysqrt{x^2+3}7xyxsqrt{x^2+3}+ysqrt{y^2+6}x^2+y^2+2end{matrix}right.left{{}begin{matrix}xsqrt{y^2+6}+ysqrt{x^2+3}7xyxsqrt{x^2+3}+ysqrt{y^2+6}x^2+y^2+2end{matrix}right.
3/left{{}begin{matrix}left(2x+y-1right)left(sqrt{x+3}+sqrt{xy}+sqrt{x}right)8sqrt{x}left(sqrt{x+3}+sqrt{xy}right)^2+xy2xleft(6-xright)end...
Đọc tiếp
Giải hệ pt
1/\(\left\{{}\begin{matrix}4x\sqrt{y+1}+8x=\left(4x^2-4x-3\right)\sqrt{x+1}\\\dfrac{x}{x+1}+x^2=\left(y+2\right)\sqrt{\left(x+1\right)\left(y+1\right)}\end{matrix}\right.\)
2/\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)
3/\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)
4/\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)
m.n giúp e mấy bài này vs ạ!!
27 tháng 7 2021 lúc 21:39
Giải hệ pt;
\(\left\{{}\begin{matrix}xy\left(x+y\right)=x^2-xy+y^2\\\dfrac{1}{x^3}+\dfrac{1}{y^3}=16\end{matrix}\right.\)
Giải hệ phương trình sau :
a, \(\left\{{}\begin{matrix}\left(x^2+1\right)\left(y^2+1\right)=10\\\left(x+y\right)\left(xy-1\right)=3\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}x^3-1=2y\\y^3-1=2x\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}2x^2+xy=3x\\2y^2+xy=3y\end{matrix}\right.\)
27 tháng 7 2021 lúc 10:24
Giải hệ pt:
\(\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)\left(x+y-2\right)=6\\x^2+y^2-2x-2y=3\end{matrix}\right.\)
giải hệ pt:\(\left\{{}\begin{matrix}x^2+2x+y^2+y=3-xy\\xy+x+2y=1\end{matrix}\right.\)
11 tháng 8 2021 lúc 16:47
Giải hệ pt:
\(\left\{{}\begin{matrix}x^2+1+xy+y^2-4y=0\\\left(x^2+1\right)\left(x+y-2\right)=y\end{matrix}\right.\)
Giải hệ pt \(\left\{{}\begin{matrix}x^2+y^2+xy+1=4y\\y\left(x+y\right)^2=2x^2+7y+2\end{matrix}\right.\)
giải hệ pt:
(1)\(\left\{{}\begin{matrix}2\text{x}+2y+2\text{x}y=10\\x^2+y^2=5\end{matrix}\right.\)
(2)\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}=3\\\sqrt{xy}=2\end{matrix}\right.\)
(3)\(\left\{{}\begin{matrix}x-y=1\\x.y=6\end{matrix}\right.\)
(4)\(\left\{{}\begin{matrix}|x|+y=3\\2|x|-y=3\end{matrix}\right.\)