Question

Chứng minh rằng: \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}-\dfrac{1}{\left(a+b\right)^2}}=\left|\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\right|\)

Answer 1

\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\dfrac{1}{\left(a+b\right)^2}-\dfrac{2}{ab}}\)

\(=\sqrt{\left(\dfrac{a+b}{ab}\right)^2-\dfrac{1}{\left(a+b\right)^2}-2\cdot\dfrac{\left(a+b\right)}{ab}\cdot\dfrac{1}{a+b}}\)

\(=\sqrt{\left(\dfrac{a+b}{ab}-\dfrac{1}{a+b}\right)^2}\)

\(=\left|\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\right|\)