Đáp án:
a) 2x-2
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{{x^2} + 2x + {x^2} - 4}}{{x + 2}} = \dfrac{{x\left( {x + 2} \right) + \left( {x - 2} \right)\left( {x + 2} \right)}}{{x + 2}}\\
= \dfrac{{\left( {x + 2} \right)\left( {x + x - 2} \right)}}{{x + 2}}\\
= 2x - 2\\
b)\dfrac{{3{x^3} + 10{x^2} - 1}}{{3x + 1}} = \dfrac{{3{x^3} + {x^2} + 9{x^2} + 3x - 3x - 1}}{{3x + 1}}\\
= \dfrac{{{x^2}\left( {3x + 1} \right) + 3x\left( {3x + 1} \right) - \left( {3x + 1} \right)}}{{3x + 1}}\\
= \dfrac{{\left( {3x + 1} \right)\left( {{x^2} + 3x - 1} \right)}}{{3x + 1}} = {x^2} + 3x - 1\\
c)9{(x + 1)^2} + (3x - 2).(3x + 2)\\
= 9\left( {{x^2} + 2x + 1} \right) + 9{x^2} - 4\\
= 18{x^2} + 18x + 5\\
d)\dfrac{{{x^3} + 3{x^2} + 5x + 10}}{{x + 1}} = \dfrac{{{x^3} + {x^2} + 2{x^2} + 2x + 3x + 3 + 7}}{{x + 1}}\\
= \dfrac{{{x^2}\left( {x + 1} \right) + 2x\left( {x + 1} \right) + 3\left( {x + 1} \right) + 7}}{{x + 1}}\\
= \dfrac{{\left( {x + 1} \right)\left( {{x^2} + 2x + 3} \right) + 7}}{{x + 1}}\\
= {x^2} + 2x + 3 + \dfrac{7}{{x + 1}}
\end{array}\)
