Đáp án:
1) P=2
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
1)P = \sqrt {4 - 2.2.\sqrt 2 + 2} .\sqrt[3]{{20 + 14\sqrt 2 }}\\
= \sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}} .\sqrt[3]{{20 + 14\sqrt 2 }}\\
= \left( {2 - \sqrt 2 } \right)\sqrt[3]{{20 + 14\sqrt 2 }}\\
{P^3} = {\left( {2 - \sqrt 2 } \right)^3}.\left( {20 + 14\sqrt 2 } \right)\\
= \left( {20 - 14\sqrt 2 } \right)\left( {20 + 14\sqrt 2 } \right)\\
= {20^2} - {\left( {14\sqrt 2 } \right)^2} = 8\\
\to P = 2\\
2)A = \dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{1}{{x + \sqrt x + 1}} = \dfrac{1}{{x + 2.\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}}}\\
= \dfrac{1}{{{{\left( {\sqrt x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}}\\
Do:{\left( {\sqrt x + \dfrac{1}{2}} \right)^2} \ge 0\forall x \ge 0\\
\to {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\to \dfrac{1}{{{{\left( {\sqrt x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}} \le \dfrac{4}{3}\\
\to Max = \dfrac{4}{3}\\
\Leftrightarrow \sqrt x + \dfrac{1}{2} = 0\left( l \right)\left( {do:\sqrt x + \dfrac{1}{2} > 0\forall x \ge 0} \right)
\end{array}\)
⇒ Không tồn tại x thỏa mãn A đạt GTLN
