Question

Ai giúp em với

Answer 1

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

Với \(x=y\Leftrightarrow VT=\dfrac{x}{x^2+1}+\dfrac{x}{x^2+1}=\dfrac{2x}{x^2+1}\)

\(VP=\dfrac{2\left(x+x\right)}{x^2+x^2+2}=\dfrac{4x}{2x^2+2}=\dfrac{2x}{x^2+1}\)

\(\Leftrightarrow VT=VP\left(đpcm\right)\)

Với \(xy=1\Leftrightarrow VT=\dfrac{x}{x^2+xy}+\dfrac{y}{y^2+xy}=\dfrac{1}{x+y}+\dfrac{1}{y+x}=\dfrac{2}{x+y}\)

\(VP=\dfrac{2\left(x+y\right)}{x^2+y^2+2xy}=\dfrac{2\left(x+y\right)}{\left(x+y\right)^2}=\dfrac{2}{x+y}\\ \Leftrightarrow VT=VP\)

Vậy ta được đpcm

Answer 2

\(x+\dfrac{1}{x}=y+\dfrac{1}{y}\Rightarrow\dfrac{x^2+1}{x}=\dfrac{y^2+1}{y}\Rightarrow\dfrac{y}{x^2+1}=\dfrac{y}{y^2+1}\)

\(VT=\dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}=\dfrac{2x+2y}{x^2+1+y^2+1}=\dfrac{2\left(x+y\right)}{x^2+y^2+2}=VP\)(vì \(\dfrac{y}{x^2+1}=\dfrac{y}{y^2+1}\))