Đáp án:
a) $1-\sqrt[]{a}+a$
b) $A≥\dfrac{3}{4}$
Giải thích các bước giải:
a) $A=\Bigg(\dfrac{1+a\sqrt[]{a}}{1+\sqrt[]{a}}-\sqrt[]{a}\Bigg)\Bigg(\dfrac{1+\sqrt[]{a}}{1-a}\Bigg)+\dfrac{a+a\sqrt[]{a}}{1+\sqrt[]{a}}$
$=\Bigg[\dfrac{(\sqrt[]{a}+1)(a-\sqrt[]{a}+1)}{\sqrt[]{a}+1}-\sqrt[]{a}\Bigg].\dfrac{1+\sqrt[]{a}}{(1+\sqrt[]{a})(1-\sqrt[]{a})}+\dfrac{a(\sqrt[]{a}+1)}{1+\sqrt[]{a}}$
$=(a-\sqrt[]{a}+1-\sqrt[]{a}).\dfrac{1}{a}+a$
$=(a-2\sqrt[]{a}+1).\dfrac{1}{1-\sqrt[]{a}}+a$
$=\dfrac{(\sqrt[]{a}-1)^2}{1-\sqrt[]{a}}+a$
$=1-\sqrt[]{a}+a$
b) $A=a-\sqrt[]{a}+1$
$=(\sqrt[]{a})^2-2.\sqrt[]{a}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}$
$=\Bigg(\sqrt[]{a}-\dfrac{1}{2}\Bigg)^2+\dfrac{3}{4}$
$≥\dfrac{3}{4}$
Vậy $A≥\dfrac{3}{4}$
