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P
3
VA
6 tháng 10 2016
Ta có: 3 + 1 = (3^2 - 1)/(3 - 1)
3^2 + 1 = (3^4 - 1)/(3^2 - 1)
3^4 + 1 = (3^8 - 1)/(3^4 - 1)
3^8 + 1 = (3^16 - 1)/(3^8 - 1)
3^16 + 1 = (3^32 - 1)/(3^16 - 1)
3^32 + 1 = (3^64 - 1)/(3^32 - 1)
(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
=(3^2 - 1)/(3 - 1).(3^4 - 1)/(3^2 - 1).(3^8 - 1)/(3^4 - 1).(3^32 - 1)/(3^16 - 1).(3^64 - 1)/(3^32 - 1)
=(3^64 - 1)/(3 - 1)
=(3^64 - 1)/2
7 tháng 10 2016
Đặt biểu thức đó là A
(3-1) A= (3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1) (3^32+1)
2 A= (3^2-1)(3^2+1)(3^4+1)..............................................
2A = (3^4-1)(3^4+1)(3^8+1) ............................
2A= (3^8-1)(3^8+1)(3^16+1) .............
2A = (3^16-10(3^16+1)(3^32+1)
2A = (3^32-1)(3^32+1)
2A= 3^64-1
A= (3^64-1) / 2
30 tháng 6 2015
2A=2+22+23+...+2101
=>2A-A=(2+22+23+...+2101)-(1+2+22+....+2100)
=>A=2101-1
PH
5
3 tháng 7 2019
Đặt A
Rút gọn: (3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
A=(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=2(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3-1)(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3^2-1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3^4-1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3^8-1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3^16-1)(3^16 + 1)(3^32 + 1)
2A=(3^32 - 1)(3^32 + 1)
2A=3^64-1
=>A=(3^64-1) /2
3 tháng 7 2019
Lời giải :
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3^{64}-1\right)\)
\(=\frac{3^{64}-1}{2}\)
18 tháng 4 2019
A = 1 + 2 + 3 + ... + 99 + 100
Tổng A có số số hạng là \(\frac{100-1}{1}+1=100\)(số hạng)
=>\(A=\frac{\left(100+1\right).100}{2}=4950\)
18 tháng 4 2019
B = 12 + 22 + 32 + ... + 992 + 1002
Câu hỏi của Ngô Hồng Thuận - Toán lớp 7 - Học toán với OnlineMath
C = 13 + 23 + 33 + ... + 993 + 1003
https://lop67.tk/hoidap/16575/ti%CC%81nh-a-1-3-2-3-3-3-100-3-v%C3%A0-b-1-3-2-3-3-3-4-3-99-3-100-3
TN
7 tháng 7 2017
\(A=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(2A=3^{32}-1\Rightarrow A=\frac{3^{32}-1}{2}\)
4 tháng 8 2016
b) \(\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x+1\right)\left(x-1\right)\)
\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1+x^3-3x\left(x^2-1\right)\)
\(=3x^3+6x-3x^3+3x\)
\(=3x\)
4 tháng 8 2016
d) \(100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+..+\left(2+1\right)\left(2-1\right)\)
\(=100+99+98+97+..+2+1\)
\(=\frac{\left(100+1\right)\cdot100}{2}=5050\)
KT
5 tháng 7 2018
Bài 1:
a) \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+...+2+1=5050\)
b) \(B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=2^{128}-1+1=2^{128}\)
c) \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=2c^2\)
Ta có:
\(\left(2^3+1\right)\left(3^3+1\right)...\left(100^3+1\right)\)
\(=\left(2+1\right)\left(4-2+1\right)\left(3+1\right)\left(9-3+1\right)...\left(100+1\right)\left(100^2-100+1\right)\)
\(=3.3.4.7...101.9901\)
\(=\left(3.4.5...101\right)\left(3.7.13...9901\right)\)
\(\left(2^3-1\right)\left(3^3-1\right)...\left(100^3-1\right)\)
\(=\left(2-1\right)\left(4+2+1\right)\left(3-1\right)\left(9+3+1\right)...\left(100-1\right)\left(100^2+100+1\right)\)
\(=1.7.2.13.3.21...99.10101\)
\(=\left(1.2.3...99\right)\left(7.13.21.10101\right)\)
=> \(\frac{\left(2^3+1\right)\left(3^3+1\right)...\left(100^3+1\right)}{\left(2^3-1\right)\left(3^3-1\right)...\left(100^3-1\right)}\)
\(=\frac{\left(3.4.5...101\right)\left(3.7.13...9901\right)}{\left(1.2.3...99\right)\left(7.13.21.10101\right)}=\frac{\left(100.101\right).3}{\left(1.2\right).10101}=\frac{30300}{20202}\)
\(=3.3.4.7...101.9901\)