a) \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\) (ĐK: \(x>0,x\ne1\))
\(P=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(P=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)
\(P=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(P=x-\sqrt{x}+1\)
b) Ta có:
\(P=x-\sqrt{x}+1\)
\(P=\left(\sqrt{x}\right)^2-2\cdot\dfrac{1}{2}\cdot\sqrt{x}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(P=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Mà: \(P=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{1}{4}\left(tm\right)\)
Vậy: \(P_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{4}\)
c) \(Q=\dfrac{2\sqrt{x}}{P}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}\)
Ta có:
\(x+1\ge2\sqrt{x}\Rightarrow x-\sqrt{x}+1\ge\sqrt{x}\)
\(\Rightarrow Q=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}\ge\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
Khi \(Q=2\Rightarrow x=1\)
Khi \(Q=0\Rightarrow x=0\)
Vậy: ...
