Đáp án:
B2:
b) \(Max = \dfrac{7}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)DK:3x - 6 \ge 0 \to x \ge 2\\
b)A = 2.2\sqrt 2 + 5.3\sqrt 2 - 16\sqrt 2 \\
= 3\sqrt 2 \\
B2:\\
a)DK:x \ge 0;x \ne 4\\
B = \left[ {\dfrac{{x\left( {\sqrt x + 2} \right) - \left( {x - 1} \right)\left( {\sqrt x - 2} \right) - \sqrt x - 6}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}} \right]:\left[ {\dfrac{{\sqrt x + 2 - \sqrt x + 2}}{{\sqrt x - 2}}} \right]\\
= \dfrac{{x\sqrt x + 2x - x\sqrt x + 2x + \sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x - 2}}{4}\\
= \dfrac{{4x + \sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x - 2}}{4}\\
= \dfrac{{4x + \sqrt x - 2}}{{4\left( {\sqrt x + 2} \right)}}\\
b)B + 2 - \sqrt x = \dfrac{{4x + \sqrt x - 2}}{{4\left( {\sqrt x + 2} \right)}} + 2 - \sqrt x \\
= \dfrac{{4x + \sqrt x - 2 + 8\sqrt x + 16 - 4x - 8\sqrt x }}{{4\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x + 14}}{{4\left( {\sqrt x + 2} \right)}} = \dfrac{{\sqrt x + 2 + 12}}{{4\left( {\sqrt x + 2} \right)}}\\
= \dfrac{1}{4} + \dfrac{3}{{\sqrt x + 2}}\\
Do:x \ge 0\\
\to \sqrt x \ge 0 \to \sqrt x + 2 \ge 2\\
\to \dfrac{3}{{\sqrt x + 2}} \le \dfrac{3}{2}\\
\to \dfrac{1}{4} + \dfrac{3}{{\sqrt x + 2}} \le \dfrac{7}{4}\\
\to Max = \dfrac{7}{4}\\
\Leftrightarrow x = 0
\end{array}\)
