Đáp án:
d) \(\left[ \begin{array}{l}
x = 10\\
x = 14
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ {0;2} \right\}\\
A = \left( {\dfrac{x}{{x - 2}} + \dfrac{{3x - 2}}{{2x - {x^2}}}} \right):\left( {\dfrac{{x + 2}}{x} + \dfrac{{4 - x}}{{x - 2}}} \right)\\
= \left[ {\dfrac{{{x^2} - 3x + 2}}{{x\left( {x - 2} \right)}}} \right]:\dfrac{{\left( {x + 2} \right)\left( {x - 2} \right) + x\left( {4 - x} \right)}}{{x\left( {x - 2} \right)}}\\
= \dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{x\left( {x - 2} \right)}}.\dfrac{{x\left( {x - 2} \right)}}{{{x^2} - 4 + 4x - {x^2}}}\\
= \dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{x\left( {x - 2} \right)}}.\dfrac{{x\left( {x - 2} \right)}}{{4x - 4}}\\
= \dfrac{{x - 2}}{4}\\
b){x^2} - 5x + 6 = 0\\
\to {x^2} - 2x - 3x + 6 = 0\\
\to x\left( {x - 2} \right) - 3\left( {x - 2} \right) = 0\\
\to \left( {x - 2} \right)\left( {x - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\left( l \right)\\
x = 3
\end{array} \right.\\
Thay:x = 3\\
\to A = \dfrac{{3 - 2}}{4} = \dfrac{1}{4}\\
c)A \in Z\\
\Leftrightarrow \dfrac{{x - 2}}{4} \in Z\\
\Leftrightarrow x - 2 \in B\left( 4 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 4\\
x - 2 = 8\\
x - 2 = 12\\
x - 2 = 16\\
...
\end{array} \right. \to \left[ \begin{array}{l}
x = 6\\
x = 10\\
x = 14\\
x = 18\\
...
\end{array} \right.\\
d){A^2} - 5A + 6 = 0\\
\to \left( {A - 2} \right)\left( {A - 3} \right) = 0\\
\to \left[ \begin{array}{l}
A = 2\\
A = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{{x - 2}}{4} = 2\\
\dfrac{{x - 2}}{4} = 3
\end{array} \right. \to \left[ \begin{array}{l}
x - 2 = 8\\
x - 2 = 12
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 10\\
x = 14
\end{array} \right.
\end{array}\)
