Question

cho biểu thức

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

a, rút gọn biểu thức

b,định a để A >3

Answer 1

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

b) đk: a khác 1, a >/ 0

A > 3 ⇔ a-1> 3 ⇔ a > 4

Kl:.......

Answer 2

\(A=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+1\right].\left[\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}-1\right]\)\(A=\left(\sqrt{a}+1\right).\left(\sqrt{a}-1\right)\)

\(A=a-1\)

b) \(A>3\Leftrightarrow a-1>3\Leftrightarrow a>4\)