`~rai~`
\(\sqrt{(a+b)^2+\left(\dfrac{2}{a}+\dfrac{2}{b}\right)^2}\\=\sqrt{(a+b)^2+\dfrac{4}{a^2}+\dfrac{8}{ab}+\dfrac{4}{b^2}}\\=\sqrt{(a+b)^2+\dfrac{4b^2}{a^2b^2}+\dfrac{8ab}{a^2b^2}+\dfrac{4a^2}{a^2b^2}}\\=\sqrt{(a+b)^2+\dfrac{4}{a^2b^2}\left(b^2+2ab+a^2\right)}\\=\sqrt{(a+b)^2+\dfrac{4}{a^2b^2}(a+b)^2}\\=\sqrt{(a+b)^2\left(1+\dfrac{4}{a^2b^2}\right)}\\=\sqrt{(a+b)^2}.\sqrt{1+\dfrac{4}{a^2b^2}}\\=|a+b|.\sqrt{1+\dfrac{4}{a^2b^2}}\\=(a+b)\sqrt{1+\dfrac{4}{a^2b^2}}.\quad\text{(nếu đề cho }a+b>0)\)
