Đáp án: $A\ge -\dfrac14$
Giải thích các bước giải:
Ta có:
$A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1$
$\to A=\dfrac{\sqrt{a}(a\sqrt{a}+1)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}(2\sqrt{a}+1)}{\sqrt{a}}+1$
$\to A=\dfrac{\sqrt{a}(\sqrt{a}+1)(a-\sqrt{a}+1)}{a-\sqrt{a}+1}-(2\sqrt{a}+1)+1$
$\to A=\sqrt{a}(\sqrt{a}+1)-2\sqrt{a}-1+1$
$\to A=a+\sqrt{a}-2\sqrt{a}-1+1$
$\to A=a-\sqrt{a}$
$\to A=a-2.\sqrt{a}.\dfrac12+\dfrac14-\dfrac14$
$\to A=(\sqrt{a}-\dfrac12)^2-\dfrac14\ge -\dfrac14$
Dấu = xảy ra khi $\sqrt{a}-\dfrac12=0\to \sqrt{a}=\dfrac12\to a=\dfrac14$
