Đáp án:
\(\begin{array}{l}
a)P = {x^2} - x\\
b)x = 2\\
c)4 > x > - 3\\
d)Min = - \dfrac{1}{4}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \left[ {\dfrac{{\left( {\sqrt x - 1} \right)\left( {x + 2\sqrt x + 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{1}{{\sqrt x + 1}}} \right].x\left( {\sqrt x - 1} \right)\\
= \left[ {\dfrac{{x + 2\sqrt x + 2}}{{\sqrt x + 1}} - \dfrac{1}{{\sqrt x + 1}}} \right].x\left( {\sqrt x - 1} \right)\\
= \dfrac{{x + 2\sqrt x + 1}}{{\sqrt x + 1}}.x\left( {\sqrt x - 1} \right)\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x + 1}}.x\left( {\sqrt x - 1} \right)\\
= x\left( {x - 1} \right)\\
= {x^2} - x\\
b)P = 2\\
\to {x^2} - x = 2\\
\to {x^2} - x - 2 = 0\\
\to \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 1\left( l \right)
\end{array} \right.\\
c)P < 12\\
\to {x^2} - x < 12\\
\to {x^2} - x - 12 < 0\\
\to \left( {x - 4} \right)\left( {x + 3} \right) < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 4 < 0\\
x + 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 4 > 0\\
x + 3 < 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
4 > x > - 3\\
\left\{ \begin{array}{l}
x > 4\\
x < - 3
\end{array} \right.\left( l \right)
\end{array} \right.\\
\to 4 > x > 0; x khác 1 \\
d)P = {x^2} - x = {x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{4}\\
= {\left( {x - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4}\\
Do:{\left( {x - \dfrac{1}{2}} \right)^2} \ge 0\forall x > 0\\
\to {\left( {x - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4} \ge - \dfrac{1}{4}\\
\to Min = - \dfrac{1}{4}\\
\Leftrightarrow x = \dfrac{1}{2}
\end{array}\)
