Question

cho biểu thức : Q = \(\dfrac{a}{\sqrt{a^2-b^2}}-\left(1+\dfrac{a}{\sqrt{a^2-}b^2}\right):\dfrac{b}{a-\sqrt{a^2-b^2}}\) với a > b > 0

a, Rút gọn Q

b, Xác định giá trị của Q khi a = 3b

Answer 1

a: \(Q=\dfrac{a}{\sqrt{a^2-b^2}}-\dfrac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}}\cdot\dfrac{a-\sqrt{a^2-b^2}}{b}\)

\(=\dfrac{ab}{b\left(\sqrt{a^2-b^2}\right)}-\dfrac{a^2-\left(a^2-b^2\right)}{b\sqrt{a^2-b^2}}\)

\(=\dfrac{ab-a^2+a^2-b^2}{b\sqrt{a^2-b^2}}=\dfrac{ab-b^2}{b\sqrt{a^2-b^2}}=\dfrac{a-b}{\sqrt{a^2-b^2}}\)

b: Khi a=3b thì \(Q=\dfrac{3b-b}{\sqrt{9b^2-b^2}}=\dfrac{2b}{\sqrt{8b^2}}=\dfrac{2b}{2\sqrt{2}\cdot b}=\dfrac{1}{\sqrt{2}}\)