Đáp án:
\(\begin{array}{l}
1,\\
P = \dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}}\\
2,\\
Phương\,\,trình\,\,vô\,\,nghiệm\\
3,\\
{P_{\min }} = 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
1,\\
P = \left( {\dfrac{1}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{x - 1}}} \right):\left( {\dfrac{{\sqrt x }}{{\sqrt x - 1}} - 1} \right)\\
= \left( {\dfrac{1}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{{{\sqrt x }^2} - {1^2}}}} \right):\dfrac{{\sqrt x - \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}\\
= \left( {\dfrac{1}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right):\dfrac{{\sqrt x - \sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right) + \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{1}{{\sqrt x - 1}}\\
= \dfrac{{2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\left( {\sqrt x - 1} \right)\\
= \dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}}\\
2,\\
P = \dfrac{3}{2}\\
\Leftrightarrow \dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}} = \dfrac{3}{2}\\
\Leftrightarrow 2.\left( {2\sqrt x + 1} \right) = 3.\left( {\sqrt x + 1} \right)\\
\Leftrightarrow 4\sqrt x + 2 = 3\sqrt x + 3\\
\Leftrightarrow \sqrt x = 1\\
\Leftrightarrow x = 1\\
x \ge 0,x \ne 1 \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
3,\\
P = \dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}} = \dfrac{{\left( {\sqrt x + 1} \right) + \sqrt x }}{{\sqrt x + 1}} = 1 + \dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
\dfrac{{\sqrt x }}{{\sqrt x + 1}} \ge 0,\,\,\,\forall x \ge 0,x \ne 1\\
\Rightarrow P = 1 + \dfrac{{\sqrt x }}{{\sqrt x + 1}} \ge 1,\,\,\,\forall x \ge 0,x \ne 1\\
\Rightarrow {P_{\min }} = 1 \Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x + 1}} = 0 \Leftrightarrow \sqrt x = 0 \Leftrightarrow x = 0\\
\Rightarrow {P_{\min }} = 1
\end{array}\)
