Toán Lớp 9: so sánh : x+ căn x+1/ căn x với 2
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Toán Lớp 9: so sánh : x+ căn x+1/ căn x với 2
Comments ( 1 )
Dkxd:x > 0\\
\dfrac{{x + \sqrt x + 1}}{{\sqrt x }} – 2\\
= \dfrac{{x + \sqrt x + 1 – 2\sqrt x }}{{\sqrt x }}\\
= \dfrac{{x – \sqrt x + 1}}{{\sqrt x }}\\
= \sqrt x – 1 + \dfrac{1}{{\sqrt x }}\\
= \sqrt x + \dfrac{1}{{\sqrt x }} – 1\\
Theo\,Co – si:\\
\sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}} = 2\\
\Leftrightarrow \sqrt x + \dfrac{1}{{\sqrt x }} – 1 \ge 2 – 1\\
\Leftrightarrow \sqrt x + \dfrac{1}{{\sqrt x }} – 1 \ge 1 > 0\\
\Leftrightarrow \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} – 2 > 0\\
\Leftrightarrow \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} > 2
\end{array}$