đk: x≠0 và a≠0; a+x>0; a-x>0;
a≠\(\pm\)x;\(\sqrt{a+x}\ne\sqrt{a-x}\)\(A=\dfrac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=\dfrac{\left(\sqrt{a+x}+\sqrt{a-x}\right)^2}{a+x-a+x}=\dfrac{a+x+2\sqrt{\left(a+x\right)\left(a-x\right)}+a-x}{2x}=\dfrac{2a+2\sqrt{a^2-x^2}}{2x}=\dfrac{a+\sqrt{a^2-x^2}}{x}=\dfrac{a+\sqrt{a^2-\dfrac{4a^2b^2}{\left(b^2+1\right)^2}}}{\dfrac{2ab}{b^2+1}}\)
\(=\dfrac{b^2+1}{2b}+\dfrac{\left(b^2+1\right)\sqrt{a^2-\dfrac{4a^2b^2}{\left(b^2+1\right)^2}}}{2ab}=\dfrac{b^2+1}{2b}+\dfrac{\sqrt{a^2\left(b^2+1\right)^2-\left(b^2+1\right)^2\cdot\dfrac{4a^2b^2}{\left(b^2+1\right)^2}}}{2ab}\)
\(=\dfrac{b^2+1}{2b}+\dfrac{\sqrt{a^2\left(b^2+1\right)^2-4a^2b^2}}{2ab}=\dfrac{b^2+1}{2b}+\dfrac{\sqrt{b^4+2b^2+1-4b^2}}{2b}=\dfrac{b^2+1+\sqrt{\left(b^2-1\right)^2}}{2b}=\dfrac{b^2+1+b^2-1}{2b}=\dfrac{2b^2}{2b}=b\)
