Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
2x{\left( {2x - 1} \right)^2} - 3x\left( {x + 3} \right)\left( {x - 3} \right) - 4x{\left( {x + 1} \right)^2}\\
 = 2x.\left( {4{x^2} - 4x + 1} \right) - 3x.\left( {{x^2} - 9} \right) - 4x.\left( {{x^2} + 2x + 1} \right)\\
 = \left( {8{x^3} - 8{x^2} + 2x} \right) - \left( {3{x^3} - 27x} \right) - \left( {4{x^3} + 8{x^2} + 4x} \right)\\
 = 8{x^3} - 8{x^2} + 2x - 3{x^3} + 27x - 4{x^3} - 8{x^2} - 4x\\
 = {x^3} - 16{x^2} + 25x\\
b,\\
{\left( {a - b + c} \right)^2} - {\left( {b - c} \right)^2} + 2ab - 2ac\\
 = \left[ {\left( {a - b + c} \right) + \left( {b - c} \right)} \right].\left[ {\left( {a - b + c} \right) - \left( {b - c} \right)} \right] + 2a.\left( {b - c} \right)\\
 = a.\left( {a - 2b + 2c} \right) + 2a\left( {b - c} \right)\\
 = a.\left[ {\left( {a - 2b + 2c} \right) + 2.\left( {b - c} \right)} \right]\\
 = a.a = {a^2}\\
c,\\
{\left( {3x + 1} \right)^2} - 2.\left( {3x + 1} \right).\left( {3x + 5} \right) + {\left( {3x + 5} \right)^2}\\
 = {\left[ {\left( {3x + 1} \right) - \left( {3x + 5} \right)} \right]^2}\\
 = {\left( { - 4} \right)^2}\\
 = 16\\
d,\\
\left( {2 + 1} \right).\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right).\left( {{2^{16}} + 1} \right)\\
 = 1.\left( {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 - 1} \right).\left( {2 + 1} \right).\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right).\left( {{2^{16}} + 1} \right)\\
 = \left( {{2^2} - 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\\
e,\\
\left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right).\left( {{3^{16}} + 1} \right).\left( {{3^{32}} + 1} \right)\\
 = \frac{1}{2}.\left( {3 - 1} \right).\left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right).\left( {{3^{16}} + 1} \right).\left( {{3^{32}} + 1} \right)\\
 = \frac{1}{2}.\left( {{3^2} - 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right).\left( {{3^{16}} + 1} \right).\left( {{3^{32}} + 1} \right)\\
 = \frac{1}{2}.\left( {{3^4} - 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right).\left( {{3^{16}} + 1} \right).\left( {{3^{32}} + 1} \right)\\
 = \frac{1}{2}.\left( {{3^8} - 1} \right).\left( {{3^8} + 1} \right).\left( {{3^{16}} + 1} \right).\left( {{3^{32}} + 1} \right)\\
 = \frac{1}{2}.\left( {{3^{16}} - 1} \right).\left( {{3^{16}} + 1} \right).\left( {{3^{32}} + 1} \right)\\
 = \frac{1}{2}.\left( {{3^{32}} - 1} \right).\left( {{3^{32}} + 1} \right)\\
 = \frac{1}{2}.\left( {{3^{64}} - 1} \right)\\
f,\\
10.\left( {11 + 1} \right).\left( {{{11}^2} + 1} \right).\left( {{{11}^4} + 1} \right).\left( {{{11}^8} + 1} \right).\left( {{{11}^{16}} + 1} \right).\left( {{{11}^{32}} + 1} \right) - {11^{64}}\\
 = \left( {11 - 1} \right).\left( {11 + 1} \right).\left( {{{11}^2} + 1} \right).\left( {{{11}^4} + 1} \right).\left( {{{11}^8} + 1} \right).\left( {{{11}^{16}} + 1} \right).\left( {{{11}^{32}} + 1} \right) - {11^{64}}\\
 = \left( {{{11}^2} - 1} \right).\left( {{{11}^2} + 1} \right).\left( {{{11}^4} + 1} \right).\left( {{{11}^8} + 1} \right).\left( {{{11}^{16}} + 1} \right).\left( {{{11}^{32}} + 1} \right) - {11^{64}}\\
 = \left( {{{11}^4} - 1} \right).\left( {{{11}^4} + 1} \right).\left( {{{11}^8} + 1} \right).\left( {{{11}^{16}} + 1} \right)\left( {{{11}^{32}} + 1} \right) - {11^{64}}\\
 = \left( {{{11}^8} - 1} \right).\left( {{{11}^8} + 1} \right).\left( {{{11}^{16}} + 1} \right)\left( {{{11}^{32}} + 1} \right) - {11^{64}}\\
 = \left( {{{11}^{16}} - 1} \right)\left( {{{11}^6} + 1} \right).\left( {{{11}^{32}} + 1} \right) - {11^{64}}\\
 = \left( {{{11}^{32}} - 1} \right).\left( {{{11}^{32}} + 1} \right) - {11^{64}}\\
 = \left( {{{11}^{64}} - 1} \right) - {11^{64}}\\
 =  - 1\\
g,\\
{\left( {a + b - c} \right)^2} + {\left( {a - b + c} \right)^2} - 2.{\left( {b - c} \right)^2}\\
 = \left[ {{{\left( {a + b - c} \right)}^2} - {{\left( {b - c} \right)}^2}} \right] + \left[ {{{\left( {a - b + c} \right)}^2} - {{\left( {b - c} \right)}^2}} \right]\\
 = \left[ {\left( {a + b - c} \right) - \left( {b - c} \right)} \right].\left[ {\left( {a + b - c} \right) + \left( {b - c} \right)} \right] + \left[ {\left( {a - b + c} \right) - \left( {b - c} \right)} \right].\left[ {\left( {a - b + c} \right) + \left( {b - c} \right)} \right]\\
 = a.\left( {a + 2b - 2c} \right) + \left( {a - 2b + 2c} \right).a\\
 = a.\left[ {\left( {a + 2b - 2c} \right) + \left( {a - 2b + 2c} \right)} \right]\\
 = a.2a\\
 = 2{a^2}\\
h,\\
\left( {x - 2} \right)\left( {{x^2} - 2x + 4} \right).\left( {x + 2} \right).\left( {{x^2} + 2x + 4} \right)\\
 = \left[ {\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)} \right].\left[ {\left( {x + 2} \right).\left( {{x^2} - 2x + 4} \right)} \right]\\
 = \left( {{x^3} - 8} \right).\left( {{x^3} + 8} \right) = {x^6} - 64
\end{array}\)

Đáp án:

a) 2x(2x-1)²-3x(x+3)(x-3)-4x(x+1)²

=2x(4x²-4x+1)-3x(x²-9)-4x(x²+2x+1)

=8x³-8x²+2x-3x³+27x-4x³-8x²-4x

=x³-16x²+25x

=x(x²-16x+25)

b) (a-b+c) ²-(b-c)²+2ab-2ac

=a²+b²+c²-2ab-2bc+2ac-b²+2bc-c²+2ab-2ac

=a²

c)(3x+1)²-2(3x+1)(3x+5)+(3x+5)²

=9x²+6x+1-2(3x+3-2)(3x+3+2)+9x²+30x+25

=18x²+36x+26-2((3x+3)²-4)

=18x²+36x+26-2(9x²+18x+9-4)

=18x²+36x+26-2(9x²+18x+5)

=18x²+36x+26-18x²-36x-10

=16

d) (2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)

=1(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)

=(2-1)(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)

=(2²-1)(2²+1)(2^4+1)(2^8+1)(2^16+1)

=(2^4-1)(2^4+1)(2^8+1)(2^16+1)

=(2^8-1)(2^8+1)(2^16+1)

=(2^16-1)(2^16+1)

=2^32-1

Giải thích các bước giải:

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