Question

Rút gọn biểu thức :

\(D=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\) với O<a\(\ne4\)

Answer 1

\(D=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\cdot\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)

\(=\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\cdot\left(\dfrac{a-4}{\sqrt{a}}\right)\)

\(=\dfrac{a-2\sqrt{a}+4-a-2\sqrt{a}-4}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}\)

\(=\dfrac{-4\sqrt{a}\cdot\left(a-4\right)}{\sqrt{a}\cdot\left(a-4\right)}=-4\)

Answer 2

\(D=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\cdot\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)

\(D=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ D=\dfrac{-8\sqrt{a}}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ D=-\dfrac{8\sqrt{a}}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.\dfrac{a-4}{\sqrt{a}}\\ D=-\dfrac{8}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)=-8\)

Vậy $D=-8$