Question

P=\(\dfrac{\left(\sqrt{a^2+a\sqrt{a^2-b^2}}-\sqrt{a^2-a\sqrt{a^2-b^2}}\right)^2}{2\sqrt{a^3b}}:\left(\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b}{a}}-2\right)\left(a>0,b>0\right)\)

a)rút gọn P

b)tính P biết a=7+\(4\sqrt{3}\) , b=7-\(4\sqrt{3}\)

Answer 1

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

\(=a^2+a\sqrt{a^2-b^2}+a^2-a\sqrt{a^2-b^2}-2\cdot\sqrt{a^4-a^2\left(a^2-b^2\right)}\)

\(=2a^2-2\cdot\sqrt{a^4-a^4+a^2b^2}=2a^2-2ab\)

\(A=\left(\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b}{a}}-2\right)\)

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

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

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

b: Khi \(a=7+4\sqrt{3};b=7-4\sqrt{3}\) thì

\(P=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{2+\sqrt{3}-2+\sqrt{3}}=\dfrac{4}{2\sqrt{3}}=\dfrac{2}{\sqrt{3}}\)