2*7*7*7
>, <, =?
7…6 | 2…5 | 7…3 | 6….6 |
7…4 | 5…7 | 3…1 | 6…7 |
7…2 | 2…7 | 7…1 | 7…7 |
Lời giải chi tiết:
7 > 6 | 2 < 5 | 7 > 3 | 6 = 6 |
7 > 4 | 5 < 7 | 3 > 1 | 6 < 7 |
7 > 2 | 2 < 7 | 7 > 1 | 7 = 7 |
7 > 6 | 2 < 5 | 7 > 3 | 6 = 6 |
7 > 4 | 5 < 7 | 3 > 1 | 6 < 7 |
7 > 2 | 2 < 7 | 7 > 1 | 7 =7 |
7 > 6 7 > 4
2 < 5 5 < 7
7 > 3 3 > 1
6 = 6 6 < 7
>; <; = ?
a) 7 x 5 ? 7 x 4 7 x 2 ? 2 x 7 7 x 8 ? 7 x 9
b) 42 : 7 ? 42 : 6 21 : 7 ? 6 : 2 56 : 7 ? 49 : 7
Thực hiện phép nhân và phép chia ở cả 2 vế và so sánh.
Em điền được kết quả như sau:
a) 7 × 5 > 7 × 4 | 7 × 2 = 2 × 7 | 7 × 8 < 7 × 9 |
b) 42 : 7 < 42 : 6 | 21 : 7 = 6 : 2 | 56 : 7 > 49 : 7 |
7 x 2 = ..... 7 x 4 = ..... 7 x 6 = ..... 7 x 3 = .....
2 x 7 = ..... 4 x 7 = ..... 6 x 7 = ..... 3 x 7 = .....
14 : 7 = ..... 28 : 7 = ..... 42 : 7 = ..... 21 : 7 = .....
14 : 2 = ..... 28 : 4 = ..... 42 : 6 = ..... 21 : 3 = .....
7 x 2 = 14 7 x 4 = 28 7 x 6 = 42 7 x 3 = 21
2 x 7 = 14 4 x 7 = 28 6 x 7 = 42 3 x 7 = 21
14 : 7 = 2 28 : 7 = 4 42 : 7 = 6 21 : 7 = 3
14 : 2 = 7 28 : 4 = 7 42 : 6 = 7 21 : 3 = 7
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: \(A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\)
\(=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)
\(=7\left(1-\dfrac{1}{10}\right)\)
\(=\dfrac{63}{10}\)
Tính:
A=1+7+7^2 +7^3+..+7^2007
B=1+4+4^2+4^3+...+4^100
C=1+3^2+3^4+3^6+3^8+...+3^100
D=7+7^3+7^5+7^7+7^9+...+7^99
E=2+2^3+2^5+2^7+2^9+...+2^9009
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(7A-A=\left(7+7^2+7^3+7^4+...+7^{2008}\right)-\left(1+7+7^2+7^3+...+7^{2007}\right)\)
\(6A=7^{2008}-1\)
\(A=\frac{7^{2008}-1}{6}\)
Tương tự, \(B=\frac{4^{101}-1}{3},C=\frac{3^{101}-1}{2}\).
\(D=7+7^3+7^5+7^7+...+7^{99}\)
\(7^2.D=7^3+7^5+7^7+7^9+...+7^{101}\)
\(\left(7^2-1\right)D=\left(7^3+7^5+7^7+7^9+...+7^{101}\right)-\left(7+7^3+7^5+7^7+...+7^{99}\right)\)
\(48D=7^{101}-7\)
\(D=\frac{7^{101}-7}{48}\)
Tương tự, \(E=\frac{2^{9011}-2}{3}\)
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B = 1+7+7^2+7^3+…..+7^100
7B = 7 . (1+7+7^2 +….+7^100 )
7B = 7 + 7^2 +7^3 +…..+7^101
Lấy 7B –B ta có :
7B – B = ( 7+ 7^2+ …+7^101 )-( 1+7+7^2+….+7^100 )
=) 6B =7^101-1
=) B = 7^101-1 / 6
Đúng đấy . Cô lớp mình cho làm nhiều lần lắm.Yên tâm
Số ?
2 + ....= 7 1 + .... = 5 7 - .... = 1
7 - ....= 4 .... + 1 = 7 7 - .... = 3
....+ 3 = 7 .....+ 2 = 7 .... - 0 = 7
2 + 5 = 7 1 + 4 = 5 7 - 6 = 1
7 - 3 = 4 6 + 1 = 7 7 - 4 = 3
4 + 3 = 7 5 + 2 = 7 7 - 0 = 7
7 7 mũ 2 7 mũ 3 7 mũ 4 7 mũ 28 7 mũ 29 7 mũ 30 chia hết 57 ........ 7 mũ 28 7 mũ 29 7 mũ 2
A=7+7^2+7^3+7^4+7^5+7^6+7^7+7^8+7^9+7^11
\(7A=7^2+7^3+...+7^{12}\)
\(\Leftrightarrow A=\dfrac{7^{12}-7}{6}\)
\(7A=7^2+7^3+...+7^{10}+7^{12}\\ 7A-A=7^2+7^3+...+7^{10}+7^{12}-7-7^2-...-7^9-7^{11}\\ 6A=7^{12}-7^{11}+7^{10}-7\\ A=\dfrac{7^{12}-7^{11}+7^{10}-7}{6}\)