Question

Answer 1

b: Ta có: \(B=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}+\dfrac{6-7\sqrt{a}}{a-4}\right)\cdot\left(\sqrt{a}+2\right)\)

\(=\dfrac{a+2\sqrt{a}+\sqrt{a}-2+6-7\sqrt{a}}{\sqrt{a}-2}\)

\(=\sqrt{a}-2\)

Answer 2

\(a,A=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\left(x\ge0;x\ne4\right)\\ A=\dfrac{6x}{\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+2\right)^2}\)