Câu a : ĐKXĐ : \(a\ge0\)
\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}+1}\)
\(=\dfrac{\left(\sqrt{a}+1\right)\left(a^2+\sqrt{a}\right)-\left(2a+\sqrt{a}\right)\left(a-\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}\)
\(=\dfrac{a^2\sqrt{a}+a+a^2+\sqrt{a}-2a^2+2a\sqrt{a}-2a-a\sqrt{a}+a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}\)
\(=\dfrac{a^2\sqrt{a}-a^2+a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}\)
\(=\dfrac{a\sqrt{a}\left(a-\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}\)
\(=\dfrac{a\sqrt{a}}{\sqrt{a}+1}\)
Câu c : Vì \(a>1\Rightarrow\left|\dfrac{a\sqrt{a}}{\sqrt{a}+1}\right|=\dfrac{a\sqrt{a}}{\sqrt{a}+1}\)
Câu d : \(\dfrac{a\sqrt{a}}{\sqrt{a}+1}=2\Leftrightarrow a\sqrt{a}-2\sqrt{a}-2=0\)
Đặt \(\sqrt{a}=x\Leftrightarrow x^3-2x-2=0\) Tự giải ra nha bạn!
Câu e : \(A=\dfrac{a\sqrt{a}}{\sqrt{a}+1}\ge\dfrac{0}{\sqrt{0}+1}=0\)
Vậy GTNN của A là 0 khi \(a=0\)
