a, $A=x-\dfrac{2x-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x\sqrt{x}+1}{x-\sqrt{x}+1}+1$
$Đkxđ: +,x≥0$
$+,\sqrt{x}-1\neq0$
$⇔\sqrt{x}\neq1⇔x\neq1$
$+,x-\sqrt{x}+1\neq0$
$⇔(\sqrt{x}-\dfrac{1}{2})^2+\dfrac{3}{4}\neq0$ (luôn đúng)
Vậy để $A$ xác định thì $x≥0$ và $x\neq1$
b, $A=x-\dfrac{2x-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x\sqrt{x}+1}{x-\sqrt{x}+1}+1$
$A=x-\dfrac{2\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}-1}+\dfrac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{x-\sqrt{x}+1}+1$
$A=x-2\sqrt{x}+\sqrt{x}+1+1$
$A=x-\sqrt{x}+2$
c, $A=x-\sqrt{x}+2$
$A=(\sqrt{x})^2-2.\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{7}{4}$
$A=(\sqrt{x}-\dfrac{1}{2})^2+\dfrac{7}{4}$
$(\sqrt{x}-\dfrac{1}{2})^2≥0∀x$
$⇔(\sqrt{x}-\dfrac{1}{2})^2+\dfrac{7}{4}≥\dfrac{7}{4}∀x$
Dấu $"="$ xảy ra khi
$\sqrt{x}-\dfrac{1}{2}=0$
$⇔\sqrt{x}=\dfrac{1}{2}$
$⇔x=\dfrac{1}{4}$
Vậy $A_{min}=\dfrac{7}{4}⇔x=\dfrac{1}{4}$
