$\frac{\sqrt{x-1}}{x}_{}$ $\leq\frac{1}{2}$ $ĐK_{}$ $x_{}$ $\geq1$ Chứng minh rằng

Đáp án:

$\begin{array}{l}
\dfrac{{\sqrt {x – 1} }}{x} \le \dfrac{1}{2}\\
 \Rightarrow 2\sqrt {x – 1}  \le x\\
 \Rightarrow 4\left( {x – 1} \right) \le {x^2}\\
 \Rightarrow {x^2} – 4x + 4 \ge 0\\
 \Rightarrow {\left( {x – 2} \right)^2} \ge 0\left( {luon\,dung\,khi:x \ge 1} \right)\\
Vay\,\dfrac{{\sqrt {x – 1} }}{x} \le \dfrac{1}{2}\,khi:x \ge 1
\end{array}$