Đáp án:
$\begin{array}{l}
a)A = {x^3} + 9{x^2} + 27x + 27\\
= {\left( {x + 3} \right)^3}\\
= {\left( { - 103 + 3} \right)^3}\\
= {\left( { - 100} \right)^3}\\
= - 1000000\\
b)B = 2\left( {{x^3} + {y^3}} \right) - 3\left( {{x^2} + {y^2}} \right)\\
= 2\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) - 3{x^2} - 3{y^2}\\
= 2\left( {{x^2} - xy + {y^2}} \right) - 3{x^2} - 3{y^2}\left( {do:x + y = 1} \right)\\
= - {x^2} - 2xy - {y^2}\\
= - {\left( {x + y} \right)^2}\\
= - 1\\
c)Do:{x^2} + {y^2} = 20\\
\Leftrightarrow {x^2} + 2xy + {y^2} - 2xy = 20\\
\Leftrightarrow {\left( {x + y} \right)^2} - 2xy = 20\\
\Leftrightarrow {2^2} - 2xy = 20\\
\Leftrightarrow xy = - 8\\
C = {x^3} + {y^3}\\
= {\left( {x + y} \right)^3} - 3xy\left( {x + y} \right)\\
= {2^3} - 3xy.2\\
= 8 - 6xy\\
= 8 - 6.\left( { - 8} \right)\\
= 56\\
d)D = \left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)...\left( {{2^{64}} + 1} \right)\\
= \left( {2 - 1} \right).\left( {2 + 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)...\left( {{2^{64}} + 1} \right)\\
= \left( {{2^2} - 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)...\left( {{2^{64}} + 1} \right)\\
= \left( {{2^4} - 1} \right)\left( {{2^4} + 1} \right)...\left( {{2^{64}} + 1} \right)\\
= \left( {{2^{64}} - 1} \right)\left( {{2^{64}} + 1} \right)\\
= {2^{128}} - 1\\
e)E = {100^2} - {99^2} + {98^2} - {97^2} + ... + {2^2} - {1^2}\\
= \left( {100 - 99} \right)\left( {100 + 99} \right) + \left( {98 - 97} \right)\left( {98 + 97} \right) + ....\\
+ \left( {2 - 1} \right)\left( {2 + 1} \right)\\
= 100 + 99 + 98 + 97 + ... + 2 + 1\\
= \dfrac{{\left( {100 + 1} \right).100}}{2}\\
= 5050
\end{array}$
