Đáp án:
\(A = \left|\dfrac{1}{x} + \dfrac{1}{y} - \dfrac{1}{x+y}\right|\)
Giải thích các bước giải:
\(\begin{array}{l}
\text{Xét hằng đẳng thức:}\\
\quad \left(\dfrac1a + \dfrac1b - \dfrac{1}{a+b}\right)^2\qquad (a,b\ne 0;\ a \ne -b)\\
= \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{(a+b)^2} + \dfrac{2}{ab} - \dfrac{2}{a(a+b)} - \dfrac{2}{b(a+b)}\\
= \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{(a+b)^2} + 2\left[\dfrac{1}{ab} - \dfrac{1}{a(a+b)} - \dfrac{1}{b(a+b)}\right]\\
= \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{(a+b)^2} + 2\left[\dfrac{1}{ab} - \dfrac{a+b}{ab(a+b)}\right]\\
= \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{(a+b)^2} + 2\left(\dfrac{1}{ab} - \dfrac{1}{ab}\right)\\
= \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{(a+b)^2}\\
\text{Dựa vào hằng đẳng thức trên, ta được:}\\
\quad A = \sqrt{\dfrac{1}{x^2 + y^2} + \dfrac{1}{(x+y)^2} + \sqrt{\dfrac{1}{x^4} + \dfrac{1}{y^4} + \dfrac{1}{(x^2 + y^2)^2}}}\\
\to A = \sqrt{\dfrac{1}{x^2 + y^2} + \dfrac{1}{(x+y)^2} + \left|\dfrac{1}{x^2} + \dfrac{1}{y^2} - \dfrac{1}{x^2 + y^2}\right|}\\
\to A = \sqrt{\dfrac{1}{x^2 + y^2} + \dfrac{1}{(x+y)^2} + \dfrac{1}{x^2} + \dfrac{1}{y^2} - \dfrac{1}{x^2 + y^2}}\\
\to A = \sqrt{\dfrac{1}{(x+y)^2} + \dfrac{1}{x^2} + \dfrac{1}{y^2}}\\
\to A = \left|\dfrac{1}{x} + \dfrac{1}{y} - \dfrac{1}{x+y}\right|
\end{array}\)
