Question

Cần cách giải bài 5,giúp mình với.

Answer 1

a) điều kiện : \(a>0;a\ne1\)

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

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

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

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

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

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

\(A=\dfrac{2\sqrt{a}}{\sqrt{a}\left(a-1\right)}=\dfrac{2}{a-1}\)

c) \(A>0\Leftrightarrow\dfrac{2}{a-1}>0\Leftrightarrow a-1>0\Leftrightarrow a>1\)

vậy \(a>1\) thì \(A>0\)

d) thay \(a=\dfrac{13}{5-2\sqrt{3}}\) vào A ta có \(A=2:\dfrac{13}{5-2\sqrt{3}}=2.\dfrac{5-2\sqrt{3}}{13}=\dfrac{10-4\sqrt{3}}{13}\)