giải hệ: a, \(\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\sqrt[]{x-1}+\sqrt[]{y-1}=2\\\frac{1}{x}+\frac{1}{y}=1\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}x\sqrt[]{x}+y\sqrt[]{y}=35\\x\sqrt[]{y}+y\sqrt[]{x}=30\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}x^2+xy+y^2=3\\x+xy+y=-1\end{matrix}\right.\)
e,\(\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\)
\(e,\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy\in\left\{2;-3\right\}\end{matrix}\right.\)
Vì \(\frac{x}{y}=2>0\Rightarrow xy>0\Rightarrow xy=2\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\left(h\right)\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)
\(a,\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\frac{1}{y}\right)^2-\frac{x}{y}=3\\\left(x+\frac{1}{y}\right)+\frac{x}{y}=3\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{y}=a\\\frac{x}{y}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-b=3\\a+b=3\end{matrix}\right.\)
Làm nốt nha
\(\left\{{}\begin{matrix}\sqrt{x-1}+\sqrt{y-1}=2\\\frac{1}{x}+\frac{1}{y}=1\end{matrix}\right.\left(x;y\ge1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y-2+2\sqrt{\left(x-1\right)\left(y-1\right)}=4\\x+y=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2\sqrt{xy-\left(x+y\right)+1}=6\\x+y=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2\sqrt{xy-xy+1}=6\\x+y=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\xy=4\end{matrix}\right.\)
Làm nốt
\(c,\left\{{}\begin{matrix}x\sqrt{x}+y\sqrt{y}=35\\x\sqrt{y}+y\sqrt{x}=30\end{matrix}\right.\)\(\left(x;y\ge0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)=35\\\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}+\sqrt{y}\right)\left[\left(\sqrt{x}+\sqrt{y}\right)^2-3\sqrt{xy}\right]=35\\\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a^2-3b\right)=35\\ab=30\end{matrix}\right.\)<I>
Với \(\left\{{}\begin{matrix}a=\sqrt{x}+\sqrt{y}\ge0\\b=\sqrt{xy}\ge0\end{matrix}\right.\)
\(< I>\Leftrightarrow\left\{{}\begin{matrix}a^3-3ab=35\\ab=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^3=125\\ab=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=6\end{matrix}\right.\)
Thế vô làm nốt
\(d,\left\{{}\begin{matrix}x^2+xy+y^2=3\\x+xy+y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=3\\x+y+xy=-1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\)ta được hệ
\(\left\{{}\begin{matrix}a^2-b=3\\a+b=-1\end{matrix}\right.\)
