Đáp án:

B3:

\(\dfrac{x}{{x + 2}}\)

Giải thích các bước giải:

\(\begin{array}{l}
B1:\\
a)\dfrac{{2x - 1}}{{3{x^3}}} + \dfrac{{3x}}{{{x^3} + 1}} + \dfrac{6}{{{x^3} + 1}}\\
 = \dfrac{{\left( {{x^3} + 1} \right)\left( {2x - 1} \right) + \left( {3x + 6} \right).3{x^3}}}{{3{x^3}\left( {{x^3} + 1} \right)}}\\
 = \dfrac{{2{x^4} - {x^3} + 2x - 1 + 9{x^4} + 18{x^3}}}{{3{x^3}\left( {{x^3} + 1} \right)}}\\
 = \dfrac{{11{x^4} + 17{x^3} + 2x - 1}}{{3{x^3}\left( {{x^3} + 1} \right)}}\\
b)\dfrac{1}{{2x - 3}} - \dfrac{1}{{2x + 3}} - \dfrac{{10x + 9}}{{9 - 4{x^2}}}\\
 = \dfrac{{2x + 3 - 2x + 3 + 10x + 9}}{{\left( {2x - 3} \right)\left( {2x + 3} \right)}}\\
 = \dfrac{{10x + 15}}{{\left( {2x - 3} \right)\left( {2x + 3} \right)}} = \dfrac{{5\left( {2x + 3} \right)}}{{\left( {2x - 3} \right)\left( {2x + 3} \right)}}\\
 = \dfrac{5}{{2x - 3}}\\
B2:\\
a)4{x^3} - 4{x^2} + x = x\left( {4{x^2} - 4x + 1} \right)\\
 = x{\left( {2x - 1} \right)^2}\\
b)3x - 3y + {x^2} - {y^2}\\
 = 3\left( {x - y} \right) + \left( {x - y} \right)\left( {x + y} \right)\\
 = \left( {x - y} \right)\left( {3 + x + y} \right)\\
c){x^2} + 4xy - 4{y^2} - 1\\
 = {\left( {x - 2y} \right)^2} - 1\\
 = \left( {x - 2y + 1} \right)\left( {x - 2y - 1} \right)\\
d)16(2 - 3x) + {x^2}(3x - 2)\\
 =  - 16(3x - 2) + {x^2}(3x - 2)\\
 = \left( {3x - 2} \right)\left( {{x^2} - 16} \right)\\
 = \left( {3x - 2} \right)\left( {x - 4} \right)\left( {x + 4} \right)\\
B3:\\
\dfrac{{x - 1}}{{x + 2}} + \dfrac{{4x + 4}}{{{x^2} - 4}} + \dfrac{3}{{2 - x}}\\
 = \dfrac{{\left( {x - 1} \right)\left( {x - 2} \right) + 4x + 4 - 3\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \dfrac{{{x^2} - 3x + 2 + 4x + 4 - 3x - 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \dfrac{{{x^2} - 2x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \dfrac{x}{{x + 2}}
\end{array}\)