Cho P= $\frac{2}{x+√x+1}$
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{ "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " Cho P= $ frac{2}{x+√x+1}$ chứng minh rằng 0
Cho P= $\frac{2}{x+√x+1}$
chứng minh rằng 0
{ "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " Cho P= $ frac{2}{x+√x+1}$ chứng minh rằng 0
Đáp án:
0<P≤2
Giải thích các bước giải:
\(\begin{array}{l}
x + \sqrt x + 1 = {\left( {\sqrt x } \right)^2} + 2.\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
Do:{\left( {\sqrt x + \dfrac{1}{2}} \right)^2} > 0\forall x \ge 0\\
\to \left\{ \begin{array}{l}
{\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\\
{\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > \dfrac{3}{4}
\end{array} \right.\\
\to \dfrac{2}{{{{\left( {\sqrt x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}} < 2:\dfrac{3}{4} = \dfrac{8}{3}\\
\to P < \dfrac{8}{3}\\
KL:0 < P < \dfrac{8}{3}\\
Hay:0 < P \le 2
\end{array}\)
Dấu ”=” xảy ra
\(\begin{array}{l}
\Leftrightarrow {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} = 1\\
\Leftrightarrow {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} = \dfrac{1}{4}\\
\Leftrightarrow x + \sqrt x + \dfrac{1}{4} = \dfrac{1}{4}\\
\Leftrightarrow \sqrt x \left( {\sqrt x + 1} \right) = 0\\
\Leftrightarrow x = 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)
\end{array}\)