Question

Q=\(\dfrac{a}{\sqrt{a^2-b^2}}-\left(1+\dfrac{a}{\sqrt{a^2-b^2}}\right):\dfrac{b}{a-\sqrt{a^2-b^2}}\)

a) rút gọn Q

b) xác định giá trị của Q khi a=3b

Answer 1

a. \(Q=\dfrac{a}{\sqrt{a^2-b^2}}-\left(1+\dfrac{a}{\sqrt{a^2-b^2}}\right):\dfrac{b}{a-\sqrt{a^2-b^2}}\)

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

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

b. Thay \(a=3b\) vào \(Q\), ta được

\(Q=\dfrac{\sqrt{3b-b}}{\sqrt{3b+b}}=\dfrac{\sqrt{2b}}{\sqrt{4b}}=\dfrac{1}{\sqrt{2}}\)