Question

Answer 1

a: Ta có: \(A=\left(1+\dfrac{x+\sqrt{x}}{1+\sqrt{x}}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)

\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)

=1-x

b: Ta có: \(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\dfrac{\sqrt{x}+2}{x+16}\)

\(=\dfrac{\sqrt{x}+2}{x-16}\)