ĐKXĐ: ...
\(A=\left(\frac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\frac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\left(\frac{\sqrt{a}+2}{\sqrt{a}+1}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\right).\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}=\left(\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\frac{1}{\sqrt{a}}\)
\(=\left(\frac{a+\sqrt{a}-2-a+\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\frac{1}{\sqrt{a}}=\frac{2}{a-1}\)
\(B=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\) \(\Rightarrow B\ge0\)
\(B^2=a+2\sqrt{\frac{a^2-\left(a^2-b\right)}{4}}=a+\sqrt{b}\)
\(\Rightarrow B=\sqrt{a+\sqrt{b}}\)
