Tính :
E=7^10+7^11+7^12+......+7^30
Tính
E=7^10+7^11+7^12+....+7^30
\(E=7^{10}+7^{11}+7^{12}+...+7^{30}\)
\(7E=7^{11}+7^{12}+..+7^{31}\)
\(7E-E=\left(7^{11}+7^{12}+...+7^{31}\right)-\left(7^{10}+7^{11}+..+7^{30}\right)\)
\(6E=7^{31}-7^{10}\)
\(E=\dfrac{7^{31}-7^{10}}{6}\)
Vay...
MÌNH VAN CÁC BN GIÚP MÌNH,MÌNH CHO 10TK
Bài 1: Thực hiện phép tính (Tính hợp lý nếu có thể) a) 7/30+(-12)/37+23/30+(-25)/37 b) 5/7⋅5/11+5/7⋅2/11-5/7⋅14/11 c) (-5)/7⋅3/13-5/7⋅10/13+1 5/7
a) \(\dfrac{7}{30}+\dfrac{\left(-12\right)}{37}+\dfrac{23}{30}+\dfrac{\left(-25\right)}{37}=\left(\dfrac{7}{30}+\dfrac{23}{30}\right)+\left(\dfrac{-12}{37}+\dfrac{-25}{37}\right)=1+\left(-1\right)=0\)
b) \(\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}=\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)=-\dfrac{5}{11}\)
c) \(\dfrac{\left(-5\right)}{7}\cdot\dfrac{3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+1\dfrac{5}{7}=\dfrac{5}{7}\cdot\dfrac{-3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(\dfrac{-3}{13}-\dfrac{10}{13}\right)+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(-1\right)+\dfrac{12}{7}=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=\dfrac{7}{7}=1\)
Bài 1: A=2/3*7 + 2/7*11 + 2/11*15+ ... +2/99*103 Bài 2: A=7/2 + 7/6 + 7/12 + 7/20 + 7/30 + 7/42 + 7/56 + 7/72 + 7/90 Bài 3: A=505/10*1212 + 505/12*1414 + 505/14*1616 +...+ 505/96*9898 Bài 4: A=2/1*3 - 4/3*5 - 6/5*7 - ... - 20/19*21 Bài 5: A=1 - 5/6 + 7/12 - 9/20 + 11/30 - 13/42 + 15/56 - 17/72 + 19/90 :>
\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)
\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)
Bài 1:
Ta có: \(A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\)
\(=\dfrac{1}{2}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{100}{309}=\dfrac{50}{309}\)
Bài 2:
Ta có: \(A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\)
\(=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)
\(=7\left(1-\dfrac{1}{10}\right)\)
\(=\dfrac{63}{10}\)
hãy tính nhanh:
A.1+2+3+4+5+6+7+8+9+10+11+12+13-14=?
B.30-29+28-27+26-25+24-23+22-21...+12-11+10-9+8-7+6-5+4-3+2-1=?
Quy đồng các phân số sau:
A) 7/9 và 8/11
b) 4/5 và 7/25
c) 25/96 và 16/12
d) 1/5 , 6/10 và 12/30
e) 5/6, 7,3 và 15/24
a, \(\dfrac{7}{9}\) = \(\dfrac{7\times11}{9\times11}\) = \(\dfrac{77}{99}\)
\(\dfrac{8}{11}\) = \(\dfrac{8\times9}{11\times9}\) = \(\dfrac{72}{99}\)
Vậy \(\dfrac{7}{9}\) và \(\dfrac{8}{11}\) đã được quy đồng mẫu số lần lượt thành hai phân số:
\(\dfrac{77}{99}\) và \(\dfrac{72}{99}\)
b, \(\dfrac{4}{5}\) = \(\dfrac{4\times5}{5\times5}\) = \(\dfrac{20}{25}\)
Vậy hai phân số \(\dfrac{4}{5}\) và \(\dfrac{7}{25}\) đã được quy đồng mẫu số thành hai phân số: \(\dfrac{20}{25}\) và \(\dfrac{7}{25}\)
c, \(\dfrac{25}{96}\) và \(\dfrac{16}{12}\)
\(\dfrac{25}{96}\) = \(\dfrac{25}{96}\);
\(\dfrac{16}{12}\) = \(\dfrac{16\times8}{12\times8}\) = \(\dfrac{128}{96}\)
Vậy hai phân số \(\dfrac{25}{96}\) và \(\dfrac{16}{12}\) đã được quy đồng mẫu số thành hai phân số: \(\dfrac{25}{96}\) và \(\dfrac{128}{96}\)
Chứng mình rằng : a) (10' +8): 9 b) (1531 +2001):2 c) (10+5):3 vu9 d(11 +11 +11+. +11 +11):12 e) (7+7+7 +7) 50 (3+3+3 +3+3+3):13 { : = chia hết }
a: 10+8=18 chia hết cho 9
b: 1531 chia 2 dư 1
2001 chia 2 dư 1
=>1531+2001 chia 2 dư 2
=>1531+2001 chia hết cho 2
c: (10+5)=15 chia hết cho 3
10+5=15 ko chia hết cho 9
d:Sửa đề: 11+11^2+11^3+11^4+11^5+11^6
=11(1+11)+11^3(1+11)+11^5(1+11)
=12(11+11^3+11^5) chia hết cho 12
Tính các tổng sau bằng phương pháp hợp lí nhất:
C=3/4*7+3/7*10+3/10*13+...+3/73*76
D=7/10*11+7/11*12+7/12*13+...+7/69+70
1. Tính
a, 11/15 + - 7/10 + 13/30
b, 7/9 + -1/4 + 3/5
c, -5/21 + -5/14 + 4/35
d, 2/7 + 1/9 + 1/7 + 5/9 + 8/14
e, -4/9 + 8/15 + -2/11 + 5/-9 + 7/15
Tính:
a, 5/12+(7/59+7/12)
b,(7/30+5/16)+(1/16-7/30)
c,3/5x9+3/9x13+3/13x17+...........+3/2013-2017
d,(7/1234+51/9781-5/2018)x(3/5+1/12-37/60)
e,1/3+1/6+1/10+.........+1/4950
g,5932+6001x5931/5932x6001-69
a. 5/12+(7/59+7/12)
=5/12+497/708
=66/59
b.(7/30+5/16)+(1/16-7/30)
=131/240+(-41/240)
=3/8
a) \(\frac{5}{12}+\left(\frac{7}{59}+\frac{7}{12}\right)\)
\(=\frac{5}{12}+\frac{7}{59}+\frac{7}{12}\)
\(=\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{7}{59}\)
\(=1\frac{7}{59}\)
b) \(\left(\frac{7}{30}+\frac{5}{16}\right)+\left(\frac{1}{16}-\frac{7}{30}\right)\)
\(=\frac{7}{30}+\frac{5}{16}+\frac{1}{16}-\frac{7}{30}\)
\(=\frac{6}{16}=\frac{3}{8}\)
c) \(\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+...+\frac{3}{2013.2017}\)
\(=\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{2013}-\frac{1}{2017}\right)\)
\(=\frac{3}{4}\left(\frac{1}{5}-\frac{1}{2017}\right)\)
\(=\frac{1509}{10085}\)