Question

Cho biểu thức A = \(\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)

a, Rút gọn A

b, Tìm GTNN của A

Answer 1

ĐKXĐ: \(a>0\)

\(A=\frac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-2\sqrt{a}-1+1\)

\(=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}=a-\sqrt{a}\)

\(A=a-\sqrt{a}=a-\sqrt{a}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

\(\Rightarrow A_{min}=-\frac{1}{4}\) khi \(\sqrt{a}=\frac{1}{2}\Leftrightarrow a=\frac{1}{4}\)

Answer 2