Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
3,\\
1,\\
{\left( {6x + 1} \right)^2} + {\left( {6x - 1} \right)^2} - 2.\left( {1 + 6x} \right).\left( {6x - 1} \right)\\
= {\left( {6x + 1} \right)^2} - 2.\left( {6x + 1} \right).\left( {6x - 1} \right) + {\left( {6x - 1} \right)^2}\\
= {\left[ {\left( {6x + 1} \right) - \left( {6x - 1} \right)} \right]^2}\\
= {2^2} = 4\\
2,\\
3.\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
= \left( {{2^2} - 1} \right).\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
= \left( {{2^4} - 1} \right).\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
= \left( {{2^8} - 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
= \left( {{2^{16}} - 1} \right)\left( {{2^{16}} + 1} \right)\\
= {2^{32}} - 1\\
3,\\
x\left( {2{x^2} - 3} \right) - {x^2}\left( {5x + 1} \right) + {x^2}\\
= x.\left( {2{x^2} - 3} \right) - {x^2}.\left[ {\left( {5x + 1} \right) - 1} \right]\\
= \left( {2{x^3} - 3x} \right) - {x^2}.5x\\
= 2{x^3} - 3x - 5{x^3}\\
= - 3{x^3} - 3x\\
4.\\
3x.\left( {x - 2} \right) - 5x\left( {1 - x} \right) - 8.\left( {{x^2} - 3} \right)\\
= \left( {3{x^2} - 6x} \right) - \left( {5x - 5{x^2}} \right) - \left( {8{x^2} - 24} \right)\\
= 3{x^2} - 6x - 5x + 5{x^2} - 8{x^2} + 24\\
= - 11x + 24\\
4,\\
a,\\
{101^2} = {\left( {100 + 1} \right)^2} = {100^2} + 2.100 + 1 = 10000 + 200 + 1 = 10201\\
b,\\
97.103 = \left( {100 - 3} \right).\left( {100 + 3} \right) = {100^2} - {3^2} = 10000 - 9 = 9991\\
c,\\
{77^2} + {23^2} + 77.46 = {77^2} + 2.77.23 + {23^2} = {\left( {77 + 23} \right)^2} = {100^2} = 10000\\
d,\\
{105^2} - {5^2} = \left( {105 - 5} \right).\left( {105 + 5} \right) = 100.110 = 11000\\
e,\\
{892^2} + 892.216 + {108^2}\\
= {892^2} + 2.892.108 + {108^2}\\
= {\left( {892 + 108} \right)^2} = {1000^2} = 1000000\\
f,\\
{36^2} + {26^2} - 52.36\\
= {36^2} - 2.36.26 + {26^2}\\
= {\left( {36 - 26} \right)^2}\\
= {10^2} = 100
\end{array}\)
