a) Thay x = 25 vào A ta có:
\(A=\dfrac{25+\sqrt{25}+1}{\sqrt{25}-4}=\dfrac{25+4+1}{5-4}=30\)
b)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-8}{2\sqrt{x}-x}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-8}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{5\sqrt{x}-8}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{x-\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{x-6\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{\sqrt{x}-4}{\sqrt{x}}\)
c) \(P=A\cdot B=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-4}\cdot\dfrac{\sqrt{x}-4}{\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
Xét hiệu:
\(P-2=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-2=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}{\sqrt{x}}>0\forall x\in\left(đkxđ\right)\)
=> P > 2