Đáp án:
$\begin{array}{l}
a)\,\,\,{x^3} - 125\\
b)\,\,\,\frac{1}{2}.
\end{array}$
Giải thích các bước giải:
Câu 2:
\(\begin{array}{l}a)\,\,\,\left( {x - 5} \right)\left( {{x^2} + 26} \right) + \left( {5 - x} \right)\left( {1 - 5x} \right)\\ = \left( {x - 5} \right)\left( {{x^2} + 26} \right) + \left( {x - 5} \right)\left( {5x - 1} \right)\\ = \left( {x - 5} \right)\left( {{x^2} + 5x + 25} \right)\\ = {x^3} - {5^3} = {x^3} - 125.\\b)\,\,\left( {\frac{2}{{x - 1}} - \frac{1}{{x + 1}}} \right).\frac{{{x^2} - 1}}{{{x^2} + 6x + 9}} + \frac{{x + 1}}{{2x + 6}}\\ = \frac{{2\left( {x + 1} \right) - x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\frac{{{x^2} - 1}}{{{{\left( {x + 3} \right)}^2}}} + \frac{{x + 1}}{{2\left( {x + 3} \right)}}\\ = \frac{{2x + 2 - x + 1}}{{{{\left( {x + 3} \right)}^2}}} + \frac{{x + 1}}{{2\left( {x + 3} \right)}}\\ = \frac{{x + 3}}{{{{\left( {x + 3} \right)}^2}}} + \frac{{x + 1}}{{2\left( {x + 3} \right)}}\\ = \frac{1}{{x + 3}} + \frac{{x + 1}}{{2\left( {x + 3} \right)}}\\ = \frac{{2 + x + 1}}{{2\left( {x + 3} \right)}} = \frac{{x + 3}}{{2\left( {x + 3} \right)}} = \frac{1}{2}\end{array}\)
