`~rai~`
\(a)\text{Với }x\ge 0;x\ne 1;x\ne 4,\text{ta có:}\\P=\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\right)\\\quad=\left[\dfrac{\sqrt{x}.(\sqrt{x}+1)}{\sqrt{x}+1}-\dfrac{x+2}{\sqrt{x}+1}\right]:\left[\dfrac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}+\dfrac{\sqrt{x}-4}{(\sqrt{x}+1)(\sqrt{x}-1)}\right]\\\quad=\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\dfrac{x-\sqrt{x}+\sqrt{x}-4}{(\sqrt{x}+1)(\sqrt{x}-1)}\\\quad=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}.\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{x-4}\\\quad=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}.\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}-2)(\sqrt{x}+2)}\\\quad=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}.\\\text{Vậy với }x\ge 0;x\ne1;x\ne 4\text{ thì B=}\dfrac{\sqrt{x}-1}{\sqrt{x}+2}.\\b)\text{Thay }x=25\text{(thỏa mãn)}\text{ vào biểu thức P được:}\\P=\dfrac{\sqrt{25}-1}{\sqrt{25}+2}=\dfrac{5-1}{5+2}=\dfrac{4}{7}.\\\text{Vậy với }x=25\text{ thì giá trị biểu thức P là }\dfrac{4}{7}.\\c)P<0\\\Leftrightarrow \dfrac{\sqrt{x}-1}{\sqrt{x}+2}<0\\\Leftrightarrow \sqrt{x}-1<0\quad\text{(vì }\sqrt{x}+2\ge 2>0\quad\forall x\ge 0)\\\Leftrightarrow \sqrt{x}<1\\\Leftrightarrow (\sqrt{x})^2< 1^2\\\Leftrightarrow x< 1\\\text{Kết hợp với ĐKXĐ}\\\Rightarrow 0\le x<1\\\text{mà }x\in\mathbb{Z}\\\Rightarrow x=0.\\\text{Vậy với }x=0\text{ thì }P<0.\)
