Đáp án:
c) Min=4
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ {0;1} \right\}\\
P = \dfrac{{x\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}:\left[ {\dfrac{{\left( {x + 1} \right)\left( {x - 1} \right) + x + 2 - {x^2}}}{{x\left( {x - 1} \right)}}} \right]\\
= \dfrac{{x\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{x\left( {x - 1} \right)}}{{{x^2} - 1 + x + 2 - {x^2}}}\\
= \dfrac{{x\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{x\left( {x - 1} \right)}}{{x + 1}}\\
= \dfrac{{{x^2}}}{{x - 1}}\\
b)P < 1\\
\to \dfrac{{{x^2}}}{{x - 1}} < 1\\
\to \dfrac{{{x^2} - x + 1}}{{x - 1}} < 0\\
\to x - 1 < 0\left( {do:{x^2} - x + 1 > 0\forall x} \right)\\
\to x < 1;x \ne 0\\
c)P = \dfrac{{{x^2}}}{{x - 1}} = \dfrac{{{x^2} - 1 + 1}}{{x - 1}}\\
= \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right) + 1}}{{x - 1}}\\
= x + 1 + \dfrac{1}{{x - 1}}\\
= \left( {x - 1} \right) + \dfrac{1}{{x - 1}} + 2\\
Do:x > 1\\
BDT:Co - si:\left( {x - 1} \right) + \dfrac{1}{{x - 1}} \ge 2\sqrt {\left( {x - 1} \right).\dfrac{1}{{x - 1}}} = 2\\
\to \left( {x - 1} \right) + \dfrac{1}{{x - 1}} + 2 \ge 4\\
\to Min = 4\\
\Leftrightarrow \left( {x - 1} \right) = \dfrac{1}{{x - 1}}\\
\to x - 1 = 1\\
\to x = 2
\end{array}\)
