a: ĐKXĐ: a>0
\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\dfrac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}-1+1\)
\(=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)
b: a>1
=>\(\sqrt{a}>1\)
=>\(\sqrt{a}-1>0\)
=>\(\sqrt{a}\left(\sqrt{a}-1\right)>0\)
=>\(a-\sqrt{a}>0\)
=>A>0
=>A=|A|
c: A=2
=>\(a-\sqrt{a}=2\)
=>\(a-\sqrt{a}-2=0\)
=>\(\left(\sqrt{a}-2\right)\cdot\left(\sqrt{a}+1\right)=0\)
=>\(\sqrt{a}-2=0\)
=>a=4(nhận)
d: \(A=a-\sqrt{a}\)
\(=a-\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\)
Dấu '=' xảy ra khi \(\sqrt{a}-\dfrac{1}{2}=0\)
=>a=1/4
