Cho biểu thức A= $\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-$ $\frac{2a+\sqrt{a}}{\sqrt{a}}+1$
a.Tìm ĐKXD rồi rút gọn
b/Biết a>1 so sánh A và | A|
c/ Tìm a để A ∈Z
d/Tìm GTNN
Cho biểu thức A= $\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-$ $\frac{2a+\sqrt{a}}{\sqrt{a}}+1$
a.Tìm ĐKXD rồi rút gọn
b/Biết a>1 so sánh A và | A|
c/ Tìm a để A ∈Z
d/Tìm GTNN
….
$\begin{array}{l}A = \dfrac{a^2 + \sqrt a}{a – \sqrt a + 1} – \dfrac{2a + \sqrt a}{\sqrt a} + 1\\ a) \,\,ĐKXĐ:\,a > 0\\ A = \dfrac{\sqrt a(\sqrt{a^3} + 1)}{a – \sqrt a+ 1} – \dfrac{\sqrt a(2\sqrt a + 1)}{\sqrt a} + 1\\ \to A = \sqrt a(\sqrt a + 1) – (2\sqrt a + 1) + 1\\ \to A = a + \sqrt a – 2\sqrt a – 1 + 1\\ \to A = a – \sqrt a\\ b)\,\,A = a – \sqrt a\\ \to A = \sqrt a(\sqrt a – 1)\\ Ta\,\,có:\\ a > 1\\ \to \sqrt a > 1\\ \to \sqrt a – 1 >0\\ \to \sqrt a(\sqrt a – 1) > 0\\ \to A > 0\\ \to |A| = A\\ c)\,\,Ta\,\,có:\\ A = a – \sqrt a\\ \to A \in \Bbb Z \Leftrightarrow \sqrt a \in \Bbb Z\\ \Leftrightarrow a = k^2 \quad (k \in \Bbb N^*)\\ d)\,\,A = a – \sqrt a\\ \to A = a – 2.\dfrac{1}{2}\sqrt a + \dfrac{1}{4} – \dfrac{1}{4}\\ \to A = \left(\sqrt a – \dfrac{1}{2}\right)^2 – \dfrac{1}{4}\\ Ta\,\,có:\\ \left(\sqrt a – \dfrac{1}{2}\right)^2 \geq 0,\,\forall x >0\\ \Leftrightarrow \left(\sqrt a – \dfrac{1}{2}\right)^2 – \dfrac{1}{4} \geq – \dfrac{1}{4}\\ Hay \,\,A \geq – \dfrac{1}{4}\\ \text{Dấu = xảy ra}\,\, \Leftrightarrow \sqrt a – \dfrac{1}{2} = 0 \Leftrightarrow a = \dfrac{1}{4}\\ Vậy\,\,\min A = – \dfrac{1}{4} \Leftrightarrow a = \dfrac{1}{4}\end{array}$