a: Thay x=2 vào A, ta được:
\(A=\frac{2+\sqrt2+1}{\sqrt2+1}=\left(2+\sqrt2+1\right)\left(\sqrt2-1\right)=2\sqrt2-2+2-\sqrt2+\sqrt2-1=2-1=1\)
b: \(B=\frac{1}{\sqrt{x}-1}-\frac{x+2}{x\cdot\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{x+\sqrt{x}+1-x-2-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}-1-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(1-\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=-\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
c: C=-A*B
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}+1}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Để C là số nguyên thì \(\sqrt{x}\) ⋮\(\sqrt{x}+1\)
=>\(\sqrt{x}+1-1\) ⋮\(\sqrt{x}+1\)
=>-1⋮\(\sqrt{x}+1\)
=>\(\sqrt{x}+1=1\)
=>\(\sqrt{x}=0\)
=>x=0(nhận)
