Rút gọn biểu thức:
\(B=(2^2+1).(2^4+1).(2^8+1).(2^{16}+1)\)
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5 tháng 11 2017
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
TP
2
HA
1
20 tháng 8 2018
Đặt A=3(22 +1)(24+1)(28+1)(216+1)
=(4-1)(2^2+1)(2^4+1)(28+1)(2^16+1)
=[(2^2-1)(2^2+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)
Theo mình ý a bn làm đc
NQ
0
ND
1
M
2 tháng 11 2015
\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1-6x+1\right)^2\)
\(=2^2=4\)
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
ND
2
22 tháng 8 2015
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)\left(6x+1\right)+\left(6x-1\right)\left(6x-1\right)-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)\left(6x+1\right)+\left(6x-1\right)\left[\left(6x-1\right)-2\left(1+6x\right)\right]\)
\(=\left(6x+1\right)\left(6x+1\right)+\left(6x-1\right)\left(6x-1-2-12x\right)\)
\(=36x^2+12x+1+\left(6x-1\right)\left(-6x-3\right)\)
\(=36x^2+12x+1+\left(-36x^2-12x+3\right)\)
\(=36x^2+12x+1-36x^2-12x+3\)
\(=4\)
NM
1
KT
11 tháng 8 2018
\(A=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=> \(3A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)
=> \(A=\frac{2^{32}-1}{3}\)
\(B=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\frac{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\)
\(=\frac{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\)
\(=\frac{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\)
\(=\frac{\left(2^{16}-1\right)\left(2^{16}+1\right)}{3}\)
\(=\frac{2^{32}-1}{3}\)