Question

Cho A = \(\dfrac{a}{\sqrt{a}-1}-\dfrac{2a-\sqrt{a}}{a-\sqrt{a}} \)
a) Rút gọn
b) Tính A khi a = 3 + \(\sqrt{8}\)
c) Tìm a để A lớn hơn 0 ; A bằng 0

Answer 1

a,\(A=\dfrac{a}{\sqrt{a}-1}-\dfrac{2a-\sqrt{a}}{a-\sqrt{a}}=\dfrac{a}{\sqrt{a}-1}-\dfrac{2a-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{a\sqrt{a}-2a+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)}=\sqrt{a}-1\)

Answer 2

a, ĐKXĐ: \(x\ge0,x\ne1\)

Khi a=\(3+\sqrt{8}\)

\(\Rightarrow A=\sqrt{3+\sqrt{8}}-1=\sqrt{2+2\sqrt{2}+1}-1=\sqrt{\left(\sqrt{2}+1\right)^2}-1=\left|\sqrt{2}+1\right|-1=\sqrt{2}+1-1=\sqrt{2}\)

c,Để A>0\(\Rightarrow\sqrt{a}-1>0\Leftrightarrow\sqrt{a}>1\Leftrightarrow a>1\)

Để A=0

\(\Rightarrow A=\sqrt{a}-1=0\Leftrightarrow\sqrt{a}=1\Leftrightarrow a=1\)