Question

giải ci tiết họ e vs ạ

rút gọn

bài 5

\(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\) vs \(a>0,a\ne4\)

Answer 1

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

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

\(=\sqrt{a}\left(\sqrt{a}-2\right)\)

Answer 2

Answer 3

Q= \(\left(\dfrac{a}{\sqrt{a}.\left(\sqrt{a}-2\right)}+\dfrac{a.\sqrt{a}}{\sqrt{a}.\left(\sqrt{a}-2\right)}\right)\): \(\left(\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\right)\)
   =  \(\left(\dfrac{a+a\sqrt{a}}{\sqrt{a}.\left(\sqrt{a}-2\right)}\right)\). \(\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\sqrt{a}+1}\right)\)
   = \(\dfrac{a.\left(\sqrt{a}+1\right).\left(\sqrt{a-2}\right)^2}{\sqrt{a}\left(\sqrt{a}-2\right).\left(\sqrt{a}+1\right)}\)
   = \(a-2\sqrt{a}\)      (với a > 0 và a ≠ 4)

Answer 4

\(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\\ =\dfrac{a+a\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-2\right)}:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\\ =\dfrac{a\left(1+\sqrt{a}\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}.\dfrac{\left(\sqrt{a}-2\right)^2}{\sqrt{a}+1}\\ =\sqrt{a}\left(\sqrt{a}-2\right)\)