1) \(A=\dfrac{1+\sqrt{x}+x+\sqrt{x}}{1+\sqrt{x}}.\dfrac{1-\sqrt{x}+x-\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+1}{1+\sqrt{x}}.\dfrac{x-2\sqrt{x}+1}{1-\sqrt{x}}=\dfrac{\left(1+\sqrt{x}\right)^2}{1+\sqrt{x}}.\dfrac{\left(1-\sqrt{x}\right)^2}{1-\sqrt{x}}=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
2) \(B=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}.\dfrac{\sqrt{x}+2}{x+16}\)
\(=\dfrac{x+16}{x-16}.\dfrac{\sqrt{x}+2}{x+16}=\dfrac{\sqrt{x}+2}{x-16}\)
3) \(A=\left[\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\sqrt{a}+3}-2\right]\left[\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-1}+1\right]\)
\(=\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)=a-4\)
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: \(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\dfrac{\sqrt{x}+2}{x+16}\)
\(=\dfrac{\sqrt{x}+2}{x-16}\)
