Đáp án:
\(f)\dfrac{{ - 3{x^2}}}{{x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left( {x - {x^2} + \dfrac{1}{2}{x^3}} \right).\dfrac{2}{{3x}}\\
= \dfrac{2}{3} - \dfrac{{2x}}{3} + \dfrac{{{x^2}}}{3}\\
b)\dfrac{{x\left( {x - 1} \right) - y\left( {x - 1} \right)}}{{x\left( {x - 1} \right) + y\left( {x - 1} \right)}} + \dfrac{{2y}}{{x + y}}\\
= \dfrac{{\left( {x - 1} \right)\left( {x - y} \right)}}{{\left( {x - 1} \right)\left( {x + y} \right)}} + \dfrac{{2y}}{{x + y}}\\
= \dfrac{{x - y + 2y}}{{x + y}} = \dfrac{{x + y}}{{x + y}} = 1\\
c){\left( {x + 4} \right)^2} - 2\left( {x + 4} \right)\left( {x - 4} \right) + {\left( {x - 4} \right)^2}\\
= {\left( {x + 4 - x + 4} \right)^2} = 64\\
d)\dfrac{{3{x^3} - {x^2} - 3{x^2} + x + 12x - 4}}{{3x - 1}}\\
= \dfrac{{{x^2}\left( {3x - 1} \right) - x\left( {3x - 1} \right) + 4\left( {3x - 1} \right)}}{{3x - 1}}\\
= \dfrac{{\left( {3x - 1} \right)\left( {{x^2} - x + 4} \right)}}{{3x - 1}}\\
= {x^2} - x + 4\\
e)\dfrac{{\left( {x + 2 - 4} \right)\left( {x + 2 + 4} \right)}}{{{{\left( {x - 2} \right)}^2}}}\\
= \dfrac{{\left( {x - 2} \right)\left( {x + 6} \right)}}{{{{\left( {x - 2} \right)}^2}}}\\
= \dfrac{{x + 6}}{{x - 2}}\\
f)\dfrac{{x - 1 + 2 - 3{x^2} - 1 - x}}{{x + 1}}\\
= \dfrac{{ - 3{x^2}}}{{x + 1}}
\end{array}\)
