Đáp án:
$a)P=\dfrac{x+1}{x+\sqrt{x}+1}$
Giải thích các bước giải:
$a)P=\dfrac{x}{x+\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}\\ =\dfrac{x}{x+\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\\ =\dfrac{x}{x+\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\\ =\dfrac{x(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\dfrac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\dfrac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\\ =\dfrac{x(\sqrt{x}-1)+x+\sqrt{x}+1-x-2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\\ =\dfrac{x\sqrt{x}-x+\sqrt{x}-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\\ =\dfrac{x(\sqrt{x}-1)+\sqrt{x}-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\\ =\dfrac{x+1}{x+\sqrt{x}+1}\\ b)P>\dfrac{2}{3}(*) \ \forall \ x\ge0;x\ne 1\\ \Leftrightarrow \dfrac{x+1}{x+\sqrt{x}+1}>\dfrac{2}{3} \ \forall \ x\ge0;x\ne 1\\ \Leftrightarrow \dfrac{x+1}{x+\sqrt{x}+1}-\dfrac{2}{3}>0 \ \forall \ x\ge0;x\ne 1\\ \Leftrightarrow \dfrac{3x+3-2x-2\sqrt{x}-2}{x+\sqrt{x}+1}>0 \ \forall \ x\ge0;x\ne 1\\ \Leftrightarrow \dfrac{x-2\sqrt{x}+1}{x+\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}>0 \ \forall \ x\ge0;x\ne 1\\ \Leftrightarrow \dfrac{(\sqrt{x}-1)^2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}>0 \ \forall \ x\ge0;x\ne 1(**)$
Do các phép biến đổi là tương đương, $(**)$ đúng $\Rightarrow (*)$ đúng
Vậy $P>\dfrac{2}{3}(*) \ \forall \ x\ge0;x\ne 1.$
