Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \frac{{x - 1}}{{x + 3}} + \frac{{2x + 10}}{{x + 3}} = \frac{{x - 1 + 2x + 10}}{{x + 3}} = \frac{{3x + 9}}{{x + 3}} = 3\\
B = \frac{{4\left( {x + 3} \right)}}{{3{x^2} - x}}:\frac{{{x^2} + 3x}}{{1 - 3x}}\\
= \frac{{4\left( {x + 3} \right)}}{{x\left( {3x - 1} \right)}}.\frac{{1 - 3x}}{{x\left( {x + 3} \right)}}\\
= \frac{{4\left( {x + 3} \right)\left( {1 - 3x} \right)}}{{x\left( {3x - 1} \right)x\left( {x + 3} \right)}} = \frac{{ - 4}}{{{x^2}}}\\
C = x - {x^2} - 1 = - \left( {{x^2} - x + 1} \right) = - \left( {{x^2} - 2x.\frac{1}{2} + \frac{1}{4}} \right) - \frac{3}{4} = - {\left( {x - \frac{1}{2}} \right)^2} - \frac{3}{4} \le - \frac{3}{4} < 0,\forall x
\end{array}\)
