5-1-1
1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2=
5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5=
...................=66
...................=110
1+2+........+1+2 = 66
5+5+........+5+5 = 220
k mk nha
1/5 * 1/5 * 1/5 *1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 *
1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 *
1/5 * 1/5 =
1/5 * 1/5 * 1/5 *1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 * 1/5 = 1/522
~ Hok tốt ~
#JH
Trả lời
1/5*1/5*1/5*...*1/5*1/5=1/522
Học tốt nhé !
1+1+1+1+1+1+1+1+1+1+1+1+1+3+3+3+3+3+3+5+5+5+5+5+5+5+12547895+5547854-200
1+1+1+1+1+1+1+1+5+5+5+5+5+5+5+9+9=900
đúng hay sai
sai 100%
vì khổng thể đÂU
1 phép cộng không có số vượt qua 10 thì không thể
là số lớn hơn 100 đc
cho xin nhé
thanks
Đáp án là sai vì :
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 5 + 5 + 5 + 5+ 5 + 5 + 5 + 9 + 9
= 1 x 8 + 5 x 7 + 9 x 2
= 8 + 35 + 18
= 61
1/5+1/5^2+1/5^3+1/5^4+1/5^5+......+1/5^2012<1/4
Chứng minh rằng dãy số sau đây tăng và bị chặn trên :
\(x_1=\dfrac{1}{5+1};x_2=\dfrac{1}{5+1}+\dfrac{1}{5^2+1};x_3=\dfrac{1}{5+1}+\dfrac{1}{5^2+1}+\dfrac{1}{5^3+1},.....;x_n=\dfrac{1}{5+1}+\dfrac{1}{5^2+1}+.....+\dfrac{1}{5^n+1}\)
1+5+1+5+5+1+5+5+5+...+1+5+5+5+5+...+5+1 có 2018 Cs 5
a)(5+1/5-2/9)-(2-1/23-3/35+5/6)-(8+2/7-1/18)
b) 1/3-3/4(-3/5+1/64- -2/9-1/36+1/15
c) -5/7-(-5/67)+13/10+1/2+(-1/6)+1 3/14-(-2/5)
d)3/5:(-1/15-1/6)+3/5:(-1/3-1 1/15)
A=2 mũ 3 + 2 mũ 4 + 2 mũ 5 + 2 mũ 6 + 2 mũ 7 +.....+ 2 mũ 90
B=1+5+5 mũ 2 + 5 mũ 3 +5 mũ 4 +......+5 mũ 50
C=1/5 +1/5 mũ 2 + 1/5 mũ 3 + 1/5 mũ 4 +1/5 mũ 6 +......+1/5 mũ 102
D=1/5 +1/5 mũ 3 + 1/5 mũ 4 +1/5 mũ 5 + 1/5 mũ 6 +1/5 mũ 105
A = 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90
2A = 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100
2A - A = ( 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100 ) - ( 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 )
A = 2^100 - 2^3
B = 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50
5B = 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51
5B - B = ( 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 ) - ( 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 )
4B = 5^51 - 1
B = 5^51 - 1 / 4
Cho A=(5+1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)(5^32+1) và B=1/5(5^64-1) S^2 A và B
\(A=\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
...
\(4A=5^{64}-1\)
\(\Rightarrow A=\frac{5^{64}-1}{4}>B=\frac{5^{64}-1}{5}\)