Question

CHo:

A= \(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+4}+\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+4}\) (x ≥ 0)

a. rút gọn

b. Tính A khi x=17+\(12\sqrt{2}\)

c. Tìm x dể A=4

d. Tìm x để x>2

Answer 1

a: SỬa đề: \(A=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}\)

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

\(=x-\sqrt{x}+x+\sqrt{x}=2x\)

b: Thay \(x=17+12\sqrt{2}\) vào A, ta được:

\(A=2\left(17+12\sqrt{2}\right)=34+24\sqrt{2}\)

c: Để A=4 thì 2x=4

hay x=2