Question

Cho \(\left|x\right|\ge2,\left|y\right|\ge2\) . Chứng minh rằng phương trình \(\frac{xy}{x+y}=\frac{2003}{2004}\) vô nghiệm

Answer 1

@Akai Haruma

Answer 2

Do \(\left|x\right|\ge2;\left|y\right|\ge2\Rightarrow xy\ne0\)

Ta luôn có \(\left\{{}\begin{matrix}\frac{1}{x}\le\frac{1}{\left|x\right|}\le\frac{1}{2}\\\frac{1}{y}\le\frac{1}{\left|y\right|}\le\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\frac{1}{x}+\frac{1}{y}\le\frac{1}{2}+\frac{1}{2}=1\)

\(\frac{xy}{x+y}=\frac{2003}{2004}\Leftrightarrow\frac{x+y}{xy}=\frac{2004}{2003}\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{2004}{2003}\)

Ta có \(\frac{2004}{2003}>1\) mà \(\frac{1}{x}+\frac{1}{y}\le1\Rightarrow VT< VP\Rightarrow\) phương trình vô nghiệm