Từ pt dưới ta có: \(x^2-y.x+3=0\)
\(\Delta=y^2-12\ge0\Rightarrow y^2\ge12\) (1)
Từ pt trên ta có:
\(12-y^2=\sqrt{xy-6}\ge0\Rightarrow y^2\le12\) (2)
Từ (1); (2) \(\Rightarrow y^2=12\)
\(\Rightarrow xy=6\Rightarrow x=...\)
Từ pt dưới ta có: \(x^2-y.x+3=0\)
\(\Delta=y^2-12\ge0\Rightarrow y^2\ge12\) (1)
Từ pt trên ta có:
\(12-y^2=\sqrt{xy-6}\ge0\Rightarrow y^2\le12\) (2)
Từ (1); (2) \(\Rightarrow y^2=12\)
\(\Rightarrow xy=6\Rightarrow x=...\)
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!!
Giải hệ phương trình:
\(a,\left\{{}\begin{matrix}2x^3+x^2y+2x^2+xy+6=0\\x^2+3x+y=1\end{matrix}\right.\)
\(b,\left\{{}\begin{matrix}x^2=\left(2-y\right)\left(2+y\right)\\2x^3=\left(x+y\right)\left(4-xy\right)\end{matrix}\right.\)
\(c,\left\{{}\begin{matrix}\sqrt[3]{x+2y}=4-x-y\\\sqrt[3]{x+6}+\sqrt{2y}=2\end{matrix}\right.\)
Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\left(x+2\right)\left(y-2\right)=xy\\\left(x+4\right)\left(y-3\right)=xy+6\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\left(x+5\right)\left(y-2\right)=xy\\\left(x-5\right)\left(y+12\right)=xy\end{matrix}\right.\)
giải hệ phương trình \(\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^2+\left(\frac{x}{y}\right)^3=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\)
giải hệ phương trình: \(\left\{{}\begin{matrix}2\left(x+y\right)=3\left(\sqrt[3]{x^2y}+\sqrt[3]{xy^2}\right)\\\sqrt[3]{x}+\sqrt[3]{y}=6\end{matrix}\right.\)
giải hệ phương trình : \(\left\{{}\begin{matrix}x^3+y^3=\left(x^2+y^2\right)\sqrt{x^2-xy+y^2}\\xy=\sqrt{4x-3}\end{matrix}\right.\)
Giải hệ phương trình:
\(\left\{{}\begin{matrix}x^2+\left(y+1\right)^2=xy+x+1\\2x^3=x+y+1\end{matrix}\right.\)
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)\(\left\{{}\begin{matrix}|x|+y=3\\2|x|-y=3\end{matrix}\right.\)
giải hệ phương trình
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)\(\left\{{}\begin{matrix}x^2+y^2+2\left(xy-2\right)=0\\x^2+y^2-2xy=16\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2-2x\sqrt{y}+2y=x\\y^2-2y\sqrt{x}+2z=y\\z^2-2z\sqrt{x}+2x=z\end{matrix}\right.\)