a) điều kiện xác định : \(a>0\)
ta có : \(D=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(\Leftrightarrow D=\dfrac{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)\(\Leftrightarrow D=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b) ta có : \(D=2\Leftrightarrow x-\sqrt{x}=2\Leftrightarrow x-\sqrt{x}-2=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=4\end{matrix}\right.\)
vậy \(x=4\)
c) ta có : \(a>1\Leftrightarrow a-1>0\Leftrightarrow\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)>0\)
\(\Leftrightarrow\sqrt{a}-1>0\Leftrightarrow\sqrt{a}\left(\sqrt{a}-1\right)>0\Leftrightarrow a-\sqrt{a}>0\)
\(\Rightarrow\left|D\right|=\left|a-\sqrt{a}\right|=a-\sqrt{a}=D\) vậy \(D=\left|D\right|\)
d) ta có : \(D=a-\sqrt{a}\Leftrightarrow a-\sqrt{a}-D=0\)
phương trình này luôn có nghiệm \(\Rightarrow\Delta\ge0\)
\(\Leftrightarrow1^2-4\left(-D\right)=4D+1\ge0\Leftrightarrow D\ge\dfrac{-1}{4}\)
\(\Rightarrow D_{min}=\dfrac{-1}{4}\) khi \(\sqrt{a}=\dfrac{-b}{2a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\)
vậy \(D_{min}=\dfrac{-1}{4}\) khi \(a=\dfrac{1}{4}\)