Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix}
\frac{x+y}{xy}=2\\
(x+y)^2-2xy=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x+y=2xy\\
(x+y)^2-2xy=2\end{matrix}\right.\)
\(\Rightarrow (2xy)^2-2xy=2\)
\(\Leftrightarrow 2(xy)^2-xy-1=0\)
\(\Leftrightarrow 2xy(xy-1)+(xy-1)=0\Leftrightarrow (xy-1)(2xy+1)=0\)
\(\Leftrightarrow \left[\begin{matrix} xy=1\\ xy=\frac{-1}{2}\end{matrix}\right.\)
Nếu $xy=1\Rightarrow x+y=2xy=2$
$\Rightarrow y=2-x\Rightarrow xy=x(2-x)=1$
$\Leftrightarrow x^2-2x+1=0\Leftrightarrow (x-1)^2=0\Leftrightarrow x=1\Rightarrow y=\frac{1}{x}=1$
Nếu $xy=\frac{-1}{2}\Rightarrow x+y=2xy=-1$
$\Rightarrow y=-1-x\Rightarrow xy=x(-1-x)=\frac{-1}{2}$
$\Leftrightarrow x^2+x-\frac{1}{2}=0\Rightarrow x=\frac{-1+\sqrt{3}}{2}$
$\Rightarrow y=\frac{-1}{2x}=\frac{-1\mp \sqrt{3}}{2}$
Vậy $(x,y)=(1,1); (\frac{-1+\sqrt{3}}{2}, \frac{-1-\sqrt{3}}{2}); (\frac{-1-\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2})$
Lời giải:
HPT ⇔{x+yxy=2(x+y)2−2xy=2⇔{x+y=2xy(x+y)2−2xy=2
⇒(2xy)2−2xy=2
⇔2(xy)2−xy−1=0
⇔2xy(xy−1)+(xy−1)=0⇔(xy−1)(2xy+1)=0
⇔[xy=1xy=−12
Nếu xy=1⇒x+y=2xy=2
⇒y=2−x⇒xy=x(2−x)=1
⇔x2−2x+1=0⇔(x−1)2=0⇔x=1⇒y=1x=1
Nếu xy=−12⇒x+y=2xy=−1
⇒y=−1−x⇒xy=x(−1−x)=−12
⇔x2+x−12=0⇒x=−1+32
⇒y=−12x=−1∓32
Vậy
\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=2\\x^2+y^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{x+y}{xy}=2\\\left(x+y\right)^2-2xy=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2xy\\\left(x+y\right)^2-2xy=2\end{matrix}\right.\)
Đặt: P = xy
S = x + y ( S2 ≥ 4P )
Phương trình trở thành:
\(\left\{{}\begin{matrix}S=2P\\S^2-2P=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}S=2P\\S^2-S-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}S=2P\\\left(S-2\right)\left(S+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}S=2P\\\left[{}\begin{matrix}S=2\\S=-1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}P=1\\S=2\end{matrix}\right.\)(N) hoặc \(\left\{{}\begin{matrix}P=-\frac{1}{2}\\S=-1\end{matrix}\right.\)(N)
Tại: \(\left\{{}\begin{matrix}P=1\\S=2\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}xy=1\\x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Tại:\(\left\{{}\begin{matrix}P=-\frac{1}{2}\\S=-1\end{matrix}\right.\)
Phương trình có hai nghiệm phân biệt:
Vậy: t2 - St + P =0
\(\Leftrightarrow t^2+t-\frac{1}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{-1+\sqrt{3}}{2}\\t=\frac{-1-\sqrt{3}}{2}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\frac{-1+\sqrt{3}}{2}\\y=\frac{-1-\sqrt{3}}{2}\end{matrix}\right.\) Hoặc \(\left\{{}\begin{matrix}x=\frac{-1-\sqrt{3}}{2}\\y=\frac{-1+\sqrt{3}}{2}\end{matrix}\right.\)
Phương trình có 3 nghiệm : .........
Good luck !!!
