a)\(P=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1=\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1=\sqrt{x}\left(\sqrt{x}+1\right)-\left(2\sqrt{x}+1\right)+1=x+\sqrt{x}-2\sqrt{x}-1+1=x-\sqrt{x}\)b) Thay x=100 vào P, ta có \(P=100-\sqrt{100}=100-10=90\)
Vậy x=100 thì P=90
c) Ta có \(P=x-\sqrt{x}=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\Leftrightarrow P\ge-\dfrac{1}{4}\)
Dấu bằng xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}\)
Vậy GTNN của P là \(-\dfrac{1}{4}\) xảy ra khi \(x=\dfrac{1}{4}\)
