Question

[yuo yuo]

giải hpt: \(\left\{{}\begin{matrix}2x^2=y+\frac{1}{y}\\2y^2=x+\frac{1}{x}\end{matrix}\right.\)

Answer 1

[Trần Thanh Phương]

\(\left\{{}\begin{matrix}2x^2=y+\frac{1}{y}\left(1\right)\\2y^2=x+\frac{1}{x}\left(2\right)\end{matrix}\right.\)

Trừ theo vế 2 phương trình ta được :

\(2x^2-2y^2=y+\frac{1}{y}-x-\frac{1}{x}\)

\(\Leftrightarrow2\left(x-y\right)\left(x+y\right)+\left(x-y\right)-\frac{x-y}{xy}=0\)

\(\Leftrightarrow\left(x-y\right)\left(2x+2y+1-\frac{1}{xy}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\2x+2y+1-\frac{1}{xy}=0\end{matrix}\right.\)

+) TH1: \(x=y\)

\(\left(1\right)\Leftrightarrow2x^2=x+\frac{1}{x}\)

\(\Leftrightarrow2x^3-x^2-1=0\)

\(\Leftrightarrow2x^3-2x^2+x^2-1=0\)

\(\Leftrightarrow2x^2\left(x-1\right)+\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^2+x+1\right)=0\)

\(\Leftrightarrow x=1\)

\(\Leftrightarrow x=y=1\)

+) TH2: \(2x+2y+1-\frac{1}{xy}=0\)

Đặt \(x+y=a;xy=b\)

\(\Leftrightarrow2a+1-\frac{1}{b}=0\)

\(\Leftrightarrow2a^2b+ab-a=0\) (*)

Lấy \(\left(1\right)+\left(2\right)\Leftrightarrow2x^2+2y^2=x+y+\frac{1}{x}+\frac{1}{y}\)

\(\Leftrightarrow2\left[\left(x+y\right)^2-2xy\right]=x+y+\frac{x+y}{xy}\)

\(\Leftrightarrow2\left(a^2-b\right)=a+\frac{a}{b}\)

\(\Leftrightarrow2a^2b-4b^2=ab+a\)

\(\Leftrightarrow2a^2b+ab-a-4b^2-2ab=0\)

\(\Leftrightarrow4b^2+2ab=0\) ( theo (*) )

\(\Leftrightarrow b\left(2b+a\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}xy=0\left(3\right)\\2xy+x+y=0\left(4\right)\end{matrix}\right.\)

Vì \(x;y\ne0\) nên \(\left(3\right)\) vô nghiệm.

\(\left(4\right)\Leftrightarrow y=\frac{-x}{2x+1}\)

Khi đó \(\left(2\right)\Leftrightarrow2\cdot\left(\frac{-x}{2x+1}\right)^2=x+\frac{1}{x}\)

\(\Leftrightarrow4x^4+2x^3+5x^2+4x+1=0\)

\(\Leftrightarrow x^4+2x^3+x^2+4x^2+4x+1+3x^4=0\)

\(\Leftrightarrow\left(x^2+x\right)^2+\left(2x+1\right)^2+3x^4=0\) ( vô nghiệm )

Vậy...

Answer 2

[Nguyễn Việt Lâm]

ĐKXĐ: \(xy\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2y=y^2+1\\2xy^2=x^2+1\end{matrix}\right.\)

Chia vế cho vế ta được: \(\frac{x}{y}=\frac{y^2+1}{x^2+1}\Rightarrow x^3+x=y^3+y\)

\(\Rightarrow x^3-y^3+x-y=0\)

\(\Leftrightarrow\left(x-y\right)\left[\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1\right]=0\)

\(\Rightarrow x=y\)

Thay vào ta được: \(2x^3=x^2+1\Leftrightarrow2x^3-x^2-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^2+x+1\right)=0\)