Đáp án:
C2:
a) \(\dfrac{{1 - x}}{x}\)
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
a)P = a + b - ab\\
= 2 + \sqrt 3 + 2 - \sqrt 3 - \left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)\\
= 4 - 4 + 3 = 3\\
b)\left\{ \begin{array}{l}
3x + y = 5\\
x - 2y = - 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
6x + 2y = 10\\
x - 2y = - 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
7x = 7\\
y = 5 - 3x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y = 2
\end{array} \right.\\
C2:\\
a)P = \left[ {\dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right]:\dfrac{{\sqrt x }}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\\
= \dfrac{{1 - x}}{x}\\
b)P > \dfrac{1}{2}\\
\to \dfrac{{1 - x}}{x} > \dfrac{1}{2}\\
\to \dfrac{{2 - 2x - x}}{{2x}} > 0\\
\to \dfrac{{2 - 3x}}{x} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2 - 3x > 0\\
x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2 - 3x < 0\\
x < 0
\end{array} \right.
\end{array} \right. \to 0 < x < \dfrac{2}{3}
\end{array}\)
