Question

Cho biểu thức:

P = \(\dfrac{a\sqrt{a}}{\sqrt{a}-1}+\dfrac{1}{1-\sqrt{a}}\left(a\ge0,a\ne1\right)\)

a. Rút gọn P.

b. Tính giá trị P để \(a=\dfrac{9}{4}\)

Answer 1

a) điều kiện : \(a\ge0;a\ne1\)

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

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

b) thế \(a=\dfrac{9}{4}\) vào \(P\) ta có : \(P=\dfrac{9}{4}+\sqrt{\dfrac{9}{4}}+1=\dfrac{19}{4}\)

Answer 2

\(a.P=\dfrac{a\sqrt{a}}{\sqrt{a}-1}+\dfrac{1}{1-\sqrt{a}}=\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}-1}=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}-1}=a+\sqrt{a}+1\) \(b.a=\dfrac{9}{4}\left(TM\right)\) ⇒ \(\sqrt{a}=\dfrac{3}{2}\) , ta có :

\(P=\dfrac{9}{4}+\dfrac{3}{2}+1=\dfrac{19}{4}\)