Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
11,\\
a,\\
\left( {2a - b} \right)\left( {4{a^2} + 2ab + {b^2}} \right)\\
= 8{a^3} + 4{a^2}b + 2a{b^2} - 4{a^2}b - 2a{b^2} - {b^3}\\
= 8{a^3} - {b^3}\\
b,\\
\left( {2x - 1} \right)\left( {3x + 2} \right)\left( {3 - x} \right)\\
= \left( {6{x^2} + 4x - 3x - 2} \right)\left( {3 - x} \right)\\
= \left( {6{x^2} + x - 2} \right)\left( {3 - x} \right)\\
= 18{x^2} - 6{x^3} + 3x - {x^2} - 6 + 2x\\
= - 6{x^3} + 17{x^2} + 5x - 6\\
c,\\
\left( {1 - x} \right)\left( {1 + x + {x^2} + {x^3} + {x^4} + {x^5}} \right)\\
= \left( {1 + x + {x^2} + {x^3} + {x^4} + {x^5}} \right) - \left( {x + {x^2} + {x^3} + {x^4} + {x^5} + {x^6}} \right)\\
= 1 - {x^6}\\
12,\\
a,\\
A = \left( {x - 4} \right)\left( {x - 2} \right) - \left( {x - 1} \right)\left( {x - 3} \right)\\
= \left( {{x^2} - 2x - 4x + 8} \right) - \left( {{x^2} - 3x - x + 3} \right)\\
= \left( {{x^2} - 6x + 8} \right) - \left( {{x^2} - 4x + 3} \right)\\
= {x^2} - 6x + 8 - {x^2} + 4x - 3\\
= - 2x + 5\\
x = \dfrac{7}{4} \Rightarrow A = \dfrac{3}{2}\\
b,\\
B = \left( {a + 3} \right)\left( {9a - 8} \right) - \left( {9a - 1} \right)\left( {2 + a} \right)\\
= \left( {9{a^2} - 8a + 27a - 24} \right) - \left( {18a + 9{a^2} - 2 - a} \right)\\
= \left( {9{a^2} + 19a - 24} \right) - \left( {9{a^2} + 17a - 2} \right)\\
= 9{a^2} + 19a - 24 - 9{a^2} - 17a + 2\\
= 2a - 22\\
a = - \dfrac{7}{2} \Rightarrow B = - 29\\
c,\\
C = \left( {x + 1} \right)\left( {{x^2} - x + 1} \right) + x - \left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + 2014\\
= \left( {{x^3} - {x^2} + x + {x^2} - x + 1} \right) + x - \left( {{x^3} + {x^2} + x - {x^2} - x - 1} \right) + 2014\\
= \left( {{x^3} + 1} \right) + x - \left( {{x^3} - 1} \right) + 2014\\
= {x^3} + 1 + x - {x^3} + 1 + 2014\\
= x + 2016\\
x = - 2014 \Rightarrow C = 2
\end{array}\)
