Tính nhanh
1/3.2/7+1/3.4/7+1/3:7
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\(\dfrac{1}{7}\times\dfrac{21}{8}-\dfrac{3}{8}\times\dfrac{1}{7}-\dfrac{1}{7}\times\dfrac{2}{8}\\ =\dfrac{1}{7}\times\left(\dfrac{21}{8}-\dfrac{3}{8}-\dfrac{2}{8}\right)\\ =\dfrac{1}{7}\times\dfrac{16}{8}\\ =\dfrac{1}{7}\times2\\ =\dfrac{2}{7}\)
\(\dfrac{1}{7}\times\dfrac{21}{8}-\dfrac{3}{8}\times\dfrac{1}{7}-\dfrac{1}{7}\times\dfrac{2}{8}\)
=\(\dfrac{1}{7}\times\left(\dfrac{21}{8}-\dfrac{3}{8}-\dfrac{2}{8}\right)\)
=\(\dfrac{1}{7}\times2\)
=\(\dfrac{2}{7}\)
Bài 1:
a) Ta có: \(\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}\)
\(=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}\)
\(=\dfrac{1}{2}\)
c) Ta có: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)
Đặt A=1x3+3x5+5x7+7x9+...+99x101
6A=6x(1x3+3x5+5x7+7x9+...+99x101)
6A=1x3x6+3x5x6+5x7x6+7x9x6+...+99x101x6
6A=1x3x(5+1)+3x5x(7-1)+5x7x(9-3)+7x9x(11-5)+...+99x101x(103-97)
6A=1x3x5+1x3+3x5x7-3x5+5x7x9-3x5x7+7x9x11-5x7x9+...+99x101x103-99x101x97
6A=3+99x101x103
=>A=\(\frac{\text{3+99x101x103}}{6}\)
( 99 - 1 ) : 2 + 1 = 50 ( số )
làm bừa thui,ai tích mình mình tích lại
Số số hạng là :
Có số cặp là :
50 : 2 = 25 ( cặp )
Mỗi cặp có giá trị là :
99 - 97 = 2
Tổng dãy trên là :
25 x 2 = 50
Đáp số : 50
1 + 4 + 7 + 10 + 13 + 16 + 19 = 70
34 + 77 - 66 + 13 = 58
a) 17.13+17.42-17.35
=17.(13+42-35)
=17.20=340
b) [25.(18-42)-10]:4+6
=(25.2-10):4+6
=40:4+6=16
c) 36:32+23.22-32.3
=34+25-33
=81+32-27=86
d) B=3.42-22.3
=3.(16-4)
=3.12=36
e)20220+3.[52.10-(23-13)2]
=1+3.(250-100)
=1+450=451
g) 27.77+24.27-27
=27.(77+24-1)
=27.100=2700
h) 5.23+79:77-12020
=40+72-1
=89-1=88
i) 120:{54[50:2+(32-2.4)]}
=120:[54(25+1)]
=120:1404=10/117
\(\dfrac{7}{1^3\cdot2^3}+\dfrac{19}{2^3\cdot3^3}+\dfrac{37}{3^3\cdot4^3}+...+\dfrac{29701}{99^3\cdot100^3}\\ =\dfrac{2^3-1^3}{1^3\cdot2^3}+\dfrac{3^3-2^3}{2^3\cdot3^3}+\dfrac{4^3-3^3}{3^3\cdot4^3}+...+\dfrac{100^3-99^3}{99^3\cdot100^3}\\ =\dfrac{2^3}{1^3\cdot2^3}-\dfrac{1^3}{1^3\cdot2^3}+\dfrac{3^3}{2^3\cdot3^3}-\dfrac{2^3}{2^3\cdot3^3}+...+\dfrac{100^3}{99^3\cdot100^3}-\dfrac{99^3}{99^3\cdot100^3}\\ =\dfrac{1}{1^3}-\dfrac{1}{2^3}+\dfrac{1}{2^3}-\dfrac{1}{3^3}+...+\dfrac{1}{99^3}-\dfrac{1}{100^3}\\ =1-\dfrac{1}{100^3}< 1\)
Vậy ...
0.13427871148
\(\dfrac{1}{3}\left(\dfrac{2}{7}+\dfrac{4}{7}+\dfrac{1}{7}\right)=\dfrac{1}{3}\times1=\dfrac{1}{3}\)