Đáp án:
c. \( - 4\sqrt 3 + 9\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 1\\
P = \left[ {\dfrac{{3x + 3\sqrt x - 3 + \sqrt x + 2 + \sqrt x - 1}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}} \right].\left( {x - 1} \right)\\
= \dfrac{{3x + 5\sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}.\left( {x - 1} \right)\\
= \dfrac{{\left( {3\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \left( {3\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)\\
= 3x + 2\sqrt x - 1\\
b.\dfrac{1}{P} \in N\\
\Leftrightarrow \dfrac{1}{{3x + 2\sqrt x - 1}} \in N\\
\Leftrightarrow 3x + 2\sqrt x - 1 \in U\left( 1 \right)\\
\to 3x + 2\sqrt x - 1 = 1\\
\to \left[ \begin{array}{l}
\sqrt x = \dfrac{{ - 1 + \sqrt 7 }}{3}\\
\sqrt x = \dfrac{{ - 1 - \sqrt 7 }}{3}\left( l \right)
\end{array} \right.\\
\to x = \dfrac{{8 - 2\sqrt 7 }}{9}\left( l \right)\\
\to x \in \emptyset \\
c.Thay:x = 4 - 2\sqrt 3 \\
= 3 - 2\sqrt 3 .1 + 1\\
= {\left( {\sqrt 3 - 1} \right)^2}\\
\to P = 3\left( {4 - 2\sqrt 3 } \right) + 2\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} - 1\\
= 12 - 6\sqrt 3 + 2\left( {\sqrt 3 - 1} \right) - 1\\
= - 4\sqrt 3 + 9
\end{array}\)
