Question

Rút gọn: P=\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\)

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

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

\(=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\sqrt{a}}=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\)

Answer 2

\(P=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)     (ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\))
\(=\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}-2-\sqrt{a}-2\right)\left(\sqrt{a}-2+\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\right)\)
\(=\dfrac{-4.2\sqrt{a}}{\sqrt{a}}\)
\(=-8\)
#YM

Answer 3

ĐKXĐ {a>0,a≠4}

 

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

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

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

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

\(P=-8\)