Đáp án:
a, A = $\dfrac{2\sqrt{a}+2a+2}{\sqrt{a}}$
b, a$\neq$ 1
Giải thích các bước giải:
ĐK : a >0 ; a$\neq$ 1
a, pt ⇔ $\dfrac{(\sqrt{a}-1)(a+\sqrt{a}+1)}{\sqrt{a}(\sqrt{a}-1)}-\dfrac{(\sqrt{a}+1)(a-\sqrt{a}+1)}{\sqrt{a}( \sqrt{a}+1)}+\dfrac{a-1}{\sqrt{a}}.\dfrac{(\sqrt{a}+1)^2+(\sqrt{a}-1)^2}{a-1}$
⇔ $\dfrac{a+\sqrt{a}+1+a-\sqrt{a}+1}{\sqrt{a}}+ \dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\sqrt{a}}$ = $\dfrac{2\sqrt{a}+2a+2}{\sqrt{a}}$
b, A > 6 ⇔ $\dfrac{2\sqrt{a}+2a+2}{\sqrt{a}}$ > 6
⇔ $2\sqrt{a}+2a+2>6\sqrt{a}$ ⇔ $-2\sqrt{a}+a+1>0$
⇔ $(\sqrt{a}-1)^2>0$ ⇔ a$\neq$ 1
