Question

P = \(\dfrac{a\sqrt{a}}{\sqrt{a}-1}+\dfrac{1}{1-\sqrt{a}}\)

Rút gọn P 

Answer 1

ĐK: \(a\ge0;a\ne1\)

\(P=\dfrac{a\sqrt{a}}{\sqrt{a}-1}+\dfrac{1}{1-\sqrt{a}}\)

\(=\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}-1}\)

\(=\dfrac{a\sqrt{a}-1}{\sqrt{a}-1}\)

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

\(=a+\sqrt{a}+1\)

Answer 2

Giúp mình với 

Answer 3

Ta có: \(P=\dfrac{a\sqrt{a}}{\sqrt{a}-1}+\dfrac{1}{1-\sqrt{a}}=\dfrac{\left(\sqrt{a}\right)^3-1}{\sqrt{a}-1}=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}-1}=a+\sqrt{a}+1\)

Answer 4

\(P=\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}-1}=a+\sqrt{a}+1\)