biet 1^3+2^3+.....+10^3 = 3025
tính S = 2^3+4^3+6^3+.....+20^3
cho biết : 1^2 +2^2+3^2+...+10^2=385
tính A=3^2+6^2+9^2+...+30^2
cho biết : 1^3+2^3+3^3+...+10^3=3025
tính B=2^3+4^3+6^3+...+20^3
(mình sẽ ko tick cho nhg kết quả vd:b đó làm đc câu 1 câu 2 b đó ghi tương tự xin cảm ơn)
1)tính
S=2^2+4^2+6^2...+20^2 biet 1^2+2^2+3^2+...+10^2=385
2)tính
A=2^0+2^1+2^3+...+2^100
Bài 1:
\(S=2^2+4^2+6^2+...+20^2\)
\(=\left(1\cdot2\right)^2+\left(2\cdot2\right)^2+\left(2\cdot3\right)^2+...+\left(2\cdot10\right)^2\)
\(=1\cdot2^2+2^2\cdot2^2+2^2\cdot3^2+...+2^2\cdot10^2\)
\(=2^2\left(1+2^2+3^2+...+10^2\right)\)
\(=4\cdot385=1540\)
Bài 2:
\(A=2^0+2^1+2^2+...+2^{100}\)
\(A=1+2+2^2+...+2^{100}\)
\(2A=2\left(1+2+2^2+...+2^{100}\right)\)
\(2A=2+2^2+2^3+...+2^{101}\)
\(2A=\left(2+2^2+...+2^{101}\right)-\left(1+2+...+2^{100}\right)\)
\(A=2^{101}-1\)
Giải:
\(1.\) \(S=2^2+4^2+6^2+....+20^2\)
\(2^2=\left(1.2\right)^2\)
\(4^2=\left(2.2\right)^2\)
\(...\)
Vế dưới \(= \left(1.2\right)^2 + \left(2.2\right)^2 + ...+ \left(9.2\right)^2+ \left(10.2\right)^2\)
\(= 2^2.(1^2 + 2^2 + 3^2 + ...+ 9^2 + 10^2) \)
\(= 4. 385\)
\(= 1540\)
\(2.\)
\( 2A = 2^1 + 2^2 + 2^3 + 2^4 +...+\)\(2^{2011}\)
\(2A - A = ( 2^1 + 2^2 + 2^3+ 2^4 +...+ 2^{2011} ) - ( 1 + 2^2 + 2^3 +...+ 2^{2010} ) \)
\(\Rightarrow A = 2^{2011} - 1\)
\(S=2^2+4^2+6^2+..........+20^2\)
\(S=\left(1.2\right)^2+\left(2.2\right)^2+\left(2.3\right)^2+........+\left(2.10\right)^2\)
\(S=2^2\left(1^2+2^2+3^2+..........+10^2\right)\)
\(S=2^2.385\)
\(S=1540\)
\(A=2^0+2^1+2^2+.....+2^{100}\)
\(A=1+2+2^2+........+2^{100}\)
\(2A=2\left(1+2+2^2+.........+2^{100}\right)\)
\(2A=2+2^2+2^3+.........+2^{101}\)
\(2A-A=\left(2+2^2+2^3+......+2^{101}\right)-\left(1+2+2^2+........+2^{100}\right)\)\(A=2^{101}-1\)
tính s=2^3+4^3+6^3+...+20^3
biết 1^3+2^3+3^3+...+10^3=3025
biet 1^2+2^2+3^2+4^2+...+10^2=385 . Tinh nhanh s=2^2+4^2+6^2+...+20^2
biết : \(^{1^3+2^3+3^3+4^3+...+10^3}\)
tính S = \(2^3+4^3+6^3+...+20^3\)
Vì S lớn hơn dãy số trên 2^3 lần(mỗi thừa số lớn hơn 2^3 lần)
Nên S = 3025x2^3=3025x8=24200
Vậy S = 24200
nha
ta có : 23+43+63+...+203
= 2x(13+23+...+103)
=2 x 3025
=6050
Vậy S=23+43+...+203=6050
tim x, biet:
a,(3x-15)^7=0
b,4^2x-6=1
c,(3-x)^20:(3-x)^10=1(x khac 3)
d,(x-6)^3=(x-6)^2
=> \(3x-15=0\)
=> \(3x=0+15\)
=> \(3x=15\)
=> \(x=15:3\)
=> \(x=5\)
\(\left(3x-5\right)^7=0\)
\(\Rightarrow3x-5=0\)
\(\Rightarrow3x=5\)
\(\Rightarrow x=\frac{5}{3}\)
Cho \(1^3+2^3+3^3+...+10^3=3025\)
Tính S=\(2^3+4^3+6^3+...+20^3\)
Câu hỏi của Hoàng Thị Diễm Quỳnh - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo nhé!
S= \(2^3\)(\(1^3\)+\(2^3\)+\(3^3\)+...+\(10^3\))
S=8.3025 =24200
VẬY S=24200
S=23(13+23+33+...+103)
Mà 13+23+33+...+103=3025
=>S=23.3025
S=8.3025
S=24200.
Tính:
a) -5/7(14/5 - 7/10) : |-2/3| - 3/4(8/9 + 16/3) + 10/3(1/3 + 1/5)
b) 17/-26(1/6 - 5/3) : 17/13 - 20/3(2/5 - 1/4) + 2/3(6/5 - 9/2)
c) -8/9(9/8 - 3/2) + 5/4 : (5/2 - 15/4) - 3/4(10/9 - 8/3) : (-1/3)
d) 21/10 : (12/5 - 9/10) . (-4/7) - 3/2(1/6 - 7/12) + 1/5(3/2 - 1/4)
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
1 tim cac so nguyen x,y biet x/7=6/21 -5/y=20/28 1/2=x/12 x/8=-28/32 3/y=12/24 3/4=15/y 2 viet 3 phan so bang phan so -10/15