Question

Chứng minh rằng:

\(\dfrac{2}{xy}:(\dfrac{1}{x}-\dfrac{1}{y})^2:\dfrac{x^2+y^2}{(x-y)^2}\)=1

Answer 1

Sửa đề: \(\dfrac{2}{xy}:\left(\dfrac{1}{x}-\dfrac{1}{y}\right)^2:\dfrac{x^2+y^2}{\left(x-y\right)^2}=\dfrac{2xy}{x^2+y^2}\)

Ta có: \(\dfrac{2}{xy}:\left(\dfrac{1}{x}-\dfrac{1}{y}\right)^2:\dfrac{x^2+y^2}{\left(x-y\right)^2}\)

\(=\dfrac{2}{xy}:\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{2}{xy}\right):\dfrac{x^2+y^2}{\left(x-y\right)^2}\)

\(=\dfrac{2}{xy}:\left(\dfrac{x^2+y^2}{x^2y^2}-\dfrac{2xy}{x^2y^2}\right):\dfrac{x^2+y^2}{\left(x-y\right)^2}\)

\(=\dfrac{2}{xy}:\dfrac{x^2-2xy+y^2}{\left(xy\right)^2}:\dfrac{x^2+y^2}{\left(x-y\right)^2}\)

\(=\dfrac{2}{xy}\cdot\dfrac{\left(xy\right)^2}{\left(x-y\right)^2}:\dfrac{x^2+y^2}{\left(x-y\right)^2}\)

\(=\dfrac{2xy}{\left(x-y\right)^2}:\dfrac{x^2+y^2}{\left(x-y\right)^2}\)

\(=\dfrac{2xy}{\left(x-y\right)^2}\cdot\dfrac{\left(x-y\right)^2}{x^2+y^2}\)

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

Answer 2

Có thể đề bắt cm : VT \(\le1\)

áp dụng kq của bạn thịnh : VT = \(\dfrac{2xy}{x^2+y^2}\le\dfrac{2xy}{2xy}=1\)   (x² + y² \(\ge\) 2xy)