Question

Answer 1

1: \(A=\sqrt{20}-\sqrt{45}+\sqrt{6+2\sqrt5}\)

\(=2\sqrt5-3\sqrt5+\sqrt{\left(\sqrt5+1\right)^2}\)

\(=-\sqrt5+\sqrt5+1=1\)

2: \(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}}\)

\(B\le\frac12\)

=>\(B-\frac12\le0\)

=>\(\frac{\sqrt{x}-1}{\sqrt{x}}-\frac12\le0\)

=>\(\frac{2\sqrt{x}-2-\sqrt{x}}{2\sqrt{x}}\le0\)

=>\(\sqrt{x}-2\le0\)

=>\(\sqrt{x}\le2\)

=>0<x<=4

mà x là số nguyên và x<>1

nên x∈{2;3;4}