Đáp án:
$MaxP = - 5 \Leftrightarrow x = \dfrac{1}{9}$
Giải thích các bước giải:
b) Ta có:
$\begin{array}{l}
DKXD:x > 0;x \ne 1\\
B = \left( {\dfrac{1}{{x - \sqrt x }} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \left( {\dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}} + \dfrac{1}{{\sqrt x - 1}}} \right).\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x + 1}}\\
= \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}
\end{array}$
Vậy $B = \dfrac{{\sqrt x - 1}}{{\sqrt x }}$ khi $x>0;x\ne1$
d) Ta có:
$\begin{array}{l}
P=B - 9\sqrt x \\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }} - 9\sqrt x \\
= 1 - \dfrac{1}{{\sqrt x }} - 9\sqrt x \\
= 1 - \left( {\dfrac{1}{{\sqrt x }} + 9\sqrt x } \right)
\end{array}$
Lại có:
Áp dụng BĐT Cauchy cho hai số dương $\dfrac{1}{{\sqrt x }},9\sqrt x $ ta có:
$\begin{array}{l}
\dfrac{1}{{\sqrt x }} + 9\sqrt x \ge 2\sqrt {\dfrac{1}{{\sqrt x }}.9\sqrt x } = 6\\
\Rightarrow - \left( {\dfrac{1}{{\sqrt x }} + 9\sqrt x } \right) \le - 6\\
\Rightarrow 1 - \left( {\dfrac{1}{{\sqrt x }} + 9\sqrt x } \right) \le - 5\\
\Rightarrow P \le - 5
\end{array}$
Dấu bằng xảy ra $ \Leftrightarrow \dfrac{1}{{\sqrt x }} = 9\sqrt x \Leftrightarrow x = \dfrac{1}{9}$
Vậy $MaxP = - 5 \Leftrightarrow x = \dfrac{1}{9}$
