Question

Cho biểu thức:

E=\(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}\)-\(\dfrac{2a+\sqrt{a}}{\sqrt{a}}\) +1 với a>0

a) rút gọn E

b) Tính giá trị của E khi a=3-\(2\sqrt{2}\)

Answer 1

a) \(E=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1=\dfrac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}-1+1=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)b) Ta có a=3-\(2\sqrt{2}\) thì \(E=3-2\sqrt{2}-\sqrt{3-2\sqrt{2}}=3-2\sqrt{2}-\sqrt{2-2\sqrt{2}+1}=3-2\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}=3-2\sqrt{2}-\left|\sqrt{2}-1\right|=3-2\sqrt{2}-\left(\sqrt{2}-1\right)=3-2\sqrt{2}-\sqrt{2}+1=4-3\sqrt{2}\)