Đáp án:
f. \(Min = 0\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 1\\
P = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) - 6\sqrt x + 4 + 4\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + \sqrt x - 6\sqrt x + 4 + 4\sqrt x - 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x - \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
b.Thay:x = 6 - 2\sqrt 5 \\
= 5 - 2.\sqrt 5 .1 + 1\\
= {\left( {\sqrt 5 - 1} \right)^2}\\
P = \dfrac{{\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} }}{{\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} + 1}} = \dfrac{{\sqrt 5 - 1}}{{\sqrt 5 - 1 + 1}}\\
= \dfrac{{\sqrt 5 - 1}}{{\sqrt 5 }}\\
c.P = \dfrac{2}{3}\\
\to \dfrac{{\sqrt x }}{{\sqrt x + 1}} = \dfrac{2}{3}\\
\to 2\sqrt x + 2 = 3\sqrt x \\
\to \sqrt x = 2\\
\to x = 4\\
d.Do:P = \dfrac{{\sqrt x }}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 - 1}}{{\sqrt x + 1}}\\
= 1 - \dfrac{1}{{\sqrt x + 1}}\\
\to P < 1\\
e.P \in Z\\
\Leftrightarrow \dfrac{1}{{\sqrt x + 1}} \in Z\\
\to \sqrt x + 1 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 1\\
\sqrt x + 1 = - 1\left( l \right)
\end{array} \right.\\
\to x = 0\left( {TM} \right)\\
f.Do:\sqrt x \ge 0\forall x \ge 0\\
\to \sqrt x + 1 \ge 1\\
\to \dfrac{1}{{\sqrt x + 1}} \le 1\\
\to - \dfrac{1}{{\sqrt x + 1}} \ge - 1\\
\to 1 - \dfrac{1}{{\sqrt x + 1}} \ge 0\\
\to Min = 0\\
\Leftrightarrow x = 0
\end{array}\)
