Question

B=\(\left(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right)\left(\frac{\sqrt{x}-1}{\sqrt{x}}\right)\) tìm x \(\in N\) để B\(\ge\frac{1}{2}\)

Answer 1

ĐKXĐ: \(x>0;x\ne1\)

\(B=\left(\frac{\sqrt{x}-1+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\left(\frac{\sqrt{x}-1}{\sqrt{x}}\right)\)

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

\(B=\frac{2}{\sqrt{x}+1}\)

\(B\ge\frac{1}{2}\Rightarrow\frac{2}{\sqrt{x}+1}\ge\frac{1}{2}\Rightarrow\sqrt{x}+1\le4\)

\(\Rightarrow\sqrt{x}\le3\Rightarrow x\le9\)

Mà \(x\in N\Rightarrow x=\left\{2;3;4;5;6;7;8;9\right\}\)