Đặt \(\left\{{}\begin{matrix}\sqrt{x+\dfrac{1}{y}}=a\left(a\ge0\right)\\x+y=b\left(b\ge3\right)\end{matrix}\right.\), ta có hpt:
\(\left\{{}\begin{matrix}a+\sqrt{b-3}=3\left(1\right)\\a^2+b=8\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{b-3}=3-a\)
\(\Leftrightarrow\left\{{}\begin{matrix}3-a\ge0\\b-3=9-6a+a^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0\le a\le3\\b=a^2-6a+12\left(3\right)\end{matrix}\right.\). Thay (3) vào (2)
\(\Rightarrow a^2+a^2-6a+12=8\)
\(\Leftrightarrow2\left(a-1\right)\left(a-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\left(n\right)\)
TH1: \(a=1;b=7\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x+\dfrac{1}{y}}=1\left(4\right)\\x+y=7\end{matrix}\right.\). Thay \(x=7-y\) vào (4)
\(\Rightarrow7-y+\dfrac{1}{y}=1\)
\(\Leftrightarrow7y-y^2+1=y\)
\(\Leftrightarrow\left(y-3\right)^2-10=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=3+\sqrt{10}\\y=3-\sqrt{10}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4-\sqrt{10}\\x=4+\sqrt{10}\end{matrix}\right.\)
TH2: \(a=2;b=4\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x+\dfrac{1}{y}}=2\left(5\right)\\x+y=4\end{matrix}\right.\). Thay \(x=4-y\) vào (5)
\(\Rightarrow4-y+\dfrac{1}{y}=4\)
\(\Leftrightarrow4y-y^2+1=4y\)
\(\Leftrightarrow\left(1-y\right)\left(1+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)
Vậy . . .
