d) Để S nguyên thì \(x+\sqrt{x}+1⋮\sqrt{x}-1\)
\(\Leftrightarrow x-\sqrt{x}+2\sqrt{x}-2+3⋮\sqrt{x}-1\)
\(\Leftrightarrow3⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1;3\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;0;4\right\}\)
hay \(x\in\left\{0;4;16\right\}\)
1 /A
2 /A
3 /B
4 /D
5 /D
6/ C
7/ A
8/ A
9/ B
10 /D
a) Ta có: \(S=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)
\(=\left(\dfrac{x+\sqrt{x}+1}{x+1}\right):\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{x+1}\cdot\dfrac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b) Thay \(x=4-2\sqrt{3}\) vào S, ta được:
\(S=\dfrac{4-2\sqrt{3}+\sqrt{3}-1+1}{\sqrt{3}-1-1}\)
\(=\dfrac{\sqrt{3}-4}{2-\sqrt{3}}=\left(\sqrt{3}-4\right)\left(2+\sqrt{3}\right)\)
\(=2\sqrt{3}+3-8-4\sqrt{3}\)
\(=-2\sqrt{3}-5\)
