Question

A = \(\dfrac{1}{\sqrt[]{x}+3}+\dfrac{\sqrt[]{x}+9}{x-9}-\dfrac{1}{\sqrt[]{x}-3}\)

với x ≥ 0, x ≠ 9

a, Chứng minh A = \(\dfrac{1}{\sqrt{x}-3}\)

b, Tìm x để \(\dfrac{A}{B}=\dfrac{2}{7}\)

Answer 1

a: \(A=\dfrac{1}{\sqrt{x}+3}+\dfrac{\sqrt{x}+9}{x-9}-\dfrac{1}{\sqrt{x}-3}\)

\(=\dfrac{1}{\sqrt{x}+3}+\dfrac{\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}-3}\)

\(=\dfrac{\sqrt{x}-3+\sqrt{x}+9-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{1}{\sqrt{x}-3}\)