Câu 44:
a)
ĐK: \(x\ge0;x\ne1\)
Khi đó:
\(P=\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x^3}-1}\right):\left(\dfrac{3\left(\sqrt{x}+1\right)}{x-1}-\dfrac{2\sqrt{x}+5}{x-1}\right)\\ =\left(\dfrac{2x+1-x+\sqrt{x}}{\sqrt{x^3}-1}\right):\left(\dfrac{3\sqrt{x}+3-2\sqrt{x}-5}{x-1}\right)\\ =\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-2}{x-1}\\ =\dfrac{1}{\sqrt{x}-1}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-2}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
b)
Để P là số tự nhiên thì \(\left(\sqrt{x}+1\right)⋮\left(\sqrt{x}-2\right)\)
<=> \(\left(\sqrt{x}-2+3\right)⋮\left(\sqrt{x}-2\right)\)
<=> \(\sqrt{x}-2\inƯ\left(3\right)\)
<=> \(\sqrt{x}-2\in\left\{\pm1;\pm3\right\}\)
=> \(\sqrt{x}\in\left(3;5\right)\)
\(\Rightarrow x\in\left\{9;25\right\}\)
43:
a: \(A=\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-x+\sqrt{x}-1\right)\)
\(=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{-\left(\sqrt{x}-1\right)^2}{1}\)
\(=-\left(\sqrt{x}-1\right)\)
b: Khi \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}\) thì \(A=-\left(\dfrac{\sqrt{3}-1}{2}-1\right)=\dfrac{-\sqrt{3}+1}{2}+1=\dfrac{-\sqrt{3}+3}{2}\)
