Question

tìm x,y nếu:

\(xy\ne0\)và \(x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\)

giúp nhanh nhé :)

Answer 1

\(x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\Rightarrow x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}-4=0\Rightarrow\left(x^2-2+\frac{1}{x^2}\right)+\left(y^2-\cdot2+\frac{1}{y^2}\right)\)

\(\Rightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2\Rightarrow\left(\frac{x^2-1}{x^{ }}\right)^2+\left(\frac{y^2-1}{y}\right)^2=0\Rightarrow\hept{\begin{cases}\frac{x^2-1}{x}=0\\\frac{y^2-1}{y}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2-1=0\\\\y^2-1=0\end{cases}}}\)

<=> (x-1)(x+1)=0=>x=1 hoặc x=-1; (y-1)(y+1)=0=> y=1 hoặc y=-1