Đáp án:
\(\begin{array}{l}
a)\dfrac{x}{{x + 4}}\\
b)x > - 4\\
c)MinA = - 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{\left( {x - 2} \right)\left( {x - 3} \right) - 6 + 8x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x - 3}}{{x + 4}}\\
= \dfrac{{{x^2} - 5x + 6 - 6 + 8x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x - 3}}{{x + 4}}\\
= \dfrac{{{x^2} + 3x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x - 3}}{{x + 4}}\\
= \dfrac{x}{{x + 4}}\\
b)A < 1\\
\to \dfrac{x}{{x + 4}} < 1\\
\to \dfrac{{x - x - 4}}{{x + 4}} < 0\\
\to \dfrac{{ - 4}}{{x + 4}} < 0\\
\to x + 4 > 0\\
\to x > - 4\\
c)A = \dfrac{x}{{x + 4}} = \dfrac{{x + 4 - 4}}{{x + 4}}\\
= 1 - \dfrac{4}{{x + 4}}\\
A\min \Leftrightarrow \dfrac{4}{{x + 4}}\max \\
\Leftrightarrow \left( {x + 4} \right)\min \\
\Leftrightarrow x + 4 = 1\\
\to x = - 3\\
\to MinA = 1 - \dfrac{4}{{ - 3 + 4}} = - 3
\end{array}\)
