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A = 3(2²+1)(2^4 + 1)....(2^64 + 1) + 1
= (2²-1)(2²+1)(2^4 + 1)....(2^64 + 1) + 1
= (2^4 - 1)(2^4 + 1)....(2^64 + 1) + 1
= (2^8 - 1).(2^8 + 1)(2^16 + 1)(2^32 + 1)(2^64 + 1) + 1
= (2^16 - 1)(2^16 + 1)(2^32 + 1)(2^64 + 1) + 1
= (2^32 - 1)(2^32 + 1)(2^64 + 1) + 1
= (2^64 - 1)(2^64 + 1) + 1 = 2^128 - 1 + 1 = 2^128.
B = 3(2²+1)(2^4 + 1)....(2^64 + 1) + 1
= (2²-1)(2²+1)(2^4 + 1)....(2^64 + 1) + 1
= (2^4 - 1)(2^4 + 1)....(2^64 + 1) + 1
= (2^8 - 1).(2^8 + 1)(2^16 + 1)(2^32 + 1)(2^64 + 1) + 1
= (2^16 - 1)(2^16 + 1)(2^32 + 1)(2^64 + 1) + 1
= (2^32 - 1)(2^32 + 1)(2^64 + 1) + 1
= (2^64 - 1)(2^64 + 1) + 1 = 2^128 - 1 + 1 = 2^128.
Ta có \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)
\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)
\(=\left(3^{64}-1\right)\left(3^{64}+1\right)=\left(3^{128}-1\right)\)
\(\Rightarrow A=\frac{3^{128}-1}{2}\)
A = (22 - 1) (22 +1)(24 +1)...(264 +1) + 1 = (24 - 1)(24 +1)...(264 +1) + 1 = (28 -1)...(264 +1) + 1 = 2128 -1 + 1 = 2128
B=3.(2^2+1)(2^4+1)...(2^64+1)
=(2^2-1)(2^2+1)(2^4+1)...(2^64+1)
=(2^4-1)(2^4+1)...(2^64+1)
=(2^8-1)...(2^64+1)
.......
=(2^64-1)(2^64+1)
=2^128-1
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
1:
a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)
\(=4x^2-20x+25-4x^2-12x\)
=-32x+25
b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)
\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)
c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)
\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)
\(=\left(-3\right)^2+5\left(2x-3\right)\)
\(=9+10x-15=10x-6\)
2:
a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)
\(=9x^2-12x+4-5x^2+20x+4x-4\)
\(=4x^2+12x\)
b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)
\(=27-x^3+x^3-9x^2+27x-27\)
\(=-9x^2+27x\)
c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
Ta có : A = (3 + 1) (32 + 1) (34 + 1) ... (364 + 1)
=> 8A = (32 - 1)(32 + 1)(34 + 1)......(364 + 1)
=> 8A = (34 - 1)(34 + 1)......(364 + 1)
=> 8A = (364 - 1)(364 + 1)
=> A = \(\frac{3^{64}-1}{8}\)
