chứng minh rằng: $\sqrt[]{a^2+b^2+\frac{a^2}{(\frac{a}{b}+1)^2}}$ = |a+b-$\frac{a}{\frac{a}{b}+1}$| với mọi a,b (b khác 0)

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1. $\quad \sqrt{a^2 + b^2 + \dfrac{a^2}{\left(\dfrac ab +1\right)^2}}\qquad (b\ne 0)$

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

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

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

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

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

   $= \left|a+b -\dfrac{a}{\dfrac ab + 1}\right|$

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