a: \(S=\left(1+\frac{\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\cdot\sqrt{x}-x+\sqrt{x}-1}\right)\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right)\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\)
\(=\frac{x+\sqrt{x}+1}{x+1}\cdot\frac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b: Thay \(x=4-2\sqrt3=\left(\sqrt3-1\right)^2\) vào S, ta được:
\(S=\frac{4-2\sqrt3+\sqrt{\left(\sqrt3-1\right)^2}+1}{\sqrt{\left(\sqrt3-1\right)^2}-1}=\frac{5-2\sqrt3+\sqrt3-1}{\sqrt3-1-1}\)
\(=\frac{4-\sqrt3}{\sqrt3-2}=\frac{\left(4-\sqrt3\right)\left(\sqrt3+2\right)}{3-4}=\frac{4\sqrt3+8-3-2\sqrt3}{-1}=\frac{2\sqrt3+5}{-1}=-5-2\sqrt3\)
c: Thay x=1/2 vào S, ta được:
\(S=\left(\frac12+\frac{\sqrt2}{2}-1\right):\left(\frac{\sqrt2}{2}-1\right)=\frac{1+\sqrt2-2}{2}:\frac{\sqrt2-1}{2}=\frac{\sqrt2-1}{\sqrt2-1}=1\)
d: Để S là số nguyên thì \(x+\sqrt{x}+1\vdots\sqrt{x}-1\)
=>\(x-\sqrt{x}+2\sqrt{x}-2+3\vdots\sqrt{x}-1\)
=>\(3\vdots\sqrt{x}-1\)
=>\(\sqrt{x}-1\in\left\lbrace-1;1;3\right\rbrace\)
=>\(\sqrt{x}\in\left\lbrace0;2;4\right\rbrace\)
=>x∈{0;4;16}
