2.2.2=
chứng minh :
2^2^2^...^2 (n số 2) > 2.2.2....2 (n số 2) ^ 2.2.2....2 (n số 2) với mọi n >= 6
2.2.2..3.3
2.2.2.3.3
=(2.2.2).(3.3)
=23.32
=8.9
=72
Trả lời :72
2.2.2.3.3
=(2.2.2).(3.3)
=23.32
=8.9
=72
Trả lời :72
1.1.1+2.2.2+3.3.3+...+20.20.20
2.2.2...2=????????? (co n số 2)
2.2..2.2..2.2..2.2..2.2..2.2.2.
Tính tổng D=1.1.1+2.2.2+3.3.3+....+9.9.9
d= 1.1.1 + 2.2.2 + 3.3.3 + 4.4.4 + 5.5.5 + 6.6.6 + 7.7.7 + 8.8.8 + 9.9.9
1/2+1/2.2+1/2.2.2+1/2.2..... 50 thừa số
Đặt A = \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{2.2.2.....2}\)
= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)
=> 2A = \(2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)
= \(2\times\frac{1}{2}+2\times\frac{1}{2^2}+2\times\frac{1}{2^3}+...+2\times\frac{1}{2^{50}}\)
= \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\)
Lấy 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)
A = \(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{50}}\)
= \(1-\frac{1}{2^{50}}\)
Vậy \(\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{2.2.2.....2}\)= \(1-\frac{1}{2^{50}}\)
1/2+1/2.2+1/2.2.2+1/2.2.2.2+...+1/2.2.....2.2
kết bạn với mình nhé:
2.2.2+2=
1.2.3+4=
2.2.2+2
=4.2+2
=8+2
= 10
1.2.3+4
=2.3+4
=6+4
= 10
2.2.2+2 =10 1.2.3+4=10 kêta bạn nhé thank
2 x 2 x 2 + 2 = 2 x 3 + 2
= 6 + 2
= 8
1 x 2 x 3 + 4 = 2 x 3 + 4
= 6 + 4
= 10
mk kb rồi