Đáp án:
Giải thích các bước giải:
$\quad \sqrt{\dfrac{1}{m^2} + \dfrac{1}{n^2} + \dfrac{1}{(m+n)^2}}\qquad (m\ne 0;\ n\ne 0;\ m\ne -n)$
$= \sqrt{\dfrac{1}{m^2} + \dfrac{1}{n^2} + \dfrac{1}{(m+n)^2} + 2\cdot \dfrac{m + n - m - n}{mn(m+n)}}$
$= \sqrt{\dfrac{1}{m^2} + \dfrac{1}{n^2} + \dfrac{1}{(m+n)^2} + 2\cdot \left[\dfrac{m +n}{mn(m+n)} - \dfrac{m}{mn(m+n)} - \dfrac{n}{mn(m+n)}\right]}$
$= \sqrt{\dfrac{1}{m^2} + \dfrac{1}{n^2} + \dfrac{1}{(m+n)^2} + 2\cdot \left[\dfrac{1}{mn} - \dfrac{1}{n(m+n)} - \dfrac{1}{m(m+n)}\right]}$
$= \sqrt{\dfrac{1}{m^2} + \dfrac{1}{n^2} + \dfrac{1}{(m+n)^2} +\dfrac{2}{mn} - \dfrac{2}{n(m+n)} - \dfrac{2}{m(m+n)}}$
$= \sqrt{\left(\dfrac1m + \dfrac1n - \dfrac{1}{m+n}\right)^2}$
$=\left|\dfrac1m + \dfrac1n - \dfrac{1}{m+n}\right|$
