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\(\frac{3}{1\cdot2}+\frac{3}{2\cdot3}+\frac{3}{3\cdot4}+...+\frac{3}{2017\cdot2018}\)
Ta có : \(=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+\frac{3}{3}-\frac{3}{4}+...+\frac{3}{2017}-\frac{3}{2018}\)
\(=\frac{3}{1}-\frac{3}{2018}=\frac{6051}{2018}\)
Vậy \(\frac{3}{1\cdot2}+\frac{3}{2\cdot3}+\frac{3}{3\cdot4}+...+\frac{3}{2017\cdot2018}=\frac{6051}{2018}\)
3/1.2 + 3/2.3 + 3/3.4 + ... + 3/2017.2018
= \(3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)
= 3 . ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2017 - 1/2018 )
= 3 . ( 1 - 1/2018 )
= 3 . 2017/2018
= 6051/2018
Phần A sai nha phải là: 10 + 13 + ... + 79 + 82
đáp án:
SSH: A = ( 82 - 10 ) : 3 + 1 = 25
Tổng: ( 10 + 82 ) . 25 : 2 = 1150
a) A = 10+13 + ...+79 + 81
A = ( 79+10) x 24 : 2 + 81
A = 89 x 24 : 2 +81
A = 1068+ 81
A= 1149
b) chỗ " 1/12.20" phải là 1/12.13 chứ !
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{12}-\frac{1}{13}\)
\(=1-\frac{1}{13}\)
\(=\frac{12}{13}\)
c) \(C=\frac{3}{7}\times\frac{4}{13}+\frac{3}{7}+5\frac{4}{7}\)
\(C=\frac{3}{7}\times\left(\frac{4}{13}+1\right)+\frac{39}{7}\)
\(C=\frac{3}{7}\times\frac{17}{13}+\frac{39}{7}\)
\(C=\frac{51}{91}+\frac{39}{7}\)
\(C=\frac{558}{91}\)
Ta có: A = 1.2 + 2.3 + 3.4 + 4.5 +.....+ 98.99
=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + ..... +98.99.(100 - 97)
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 98.99.100
=> 3A = 98.99.100
=> A = 98.99.100 / 3
=> A = 323400
f,F=3. (1/2 .3 + 1/3.4 +...+ 1/99.100)
= 3. (1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +...+ 1/99 - 1/100
= 3. (1/2 - 1/100)
= 3. 49/100
= 147/100
g, G = 5/3. (3/1.4 + 3/4.7 +...+ 3/61.64)
= 5/3 . (1 - 1/4 + 1/4 - 1/7 +...+ 1/61 - 164
= 5/3 . (1-1/64)
= 5/3 . 63/64
= 105/64
f, \(F=\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)
\(\Leftrightarrow F=3\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(\Leftrightarrow F=3\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Leftrightarrow F=3\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\Leftrightarrow F=3\left(\frac{49}{100}\right)=\frac{147}{100}\)
g, \(G=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{61.64}\)
\(\Leftrightarrow G=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{61.64}\right)\)
\(\Leftrightarrow G=5.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{61}-\frac{1}{64}\right)\)
\(\Leftrightarrow G=\frac{5}{3}\left(1-\frac{1}{64}\right)\)
\(\Leftrightarrow G=\frac{5}{3}.\frac{63}{64}=\frac{105}{64}\)
\(G=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{61.64}\)
\(\Rightarrow G=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+..+\frac{3}{61.64}\right)\)
\(\Rightarrow G=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+..+\frac{1}{61}-\frac{1}{64}\right)\)
\(\Rightarrow G=\frac{5}{3}.\left(1-\frac{1}{64}\right)=\frac{5}{3}.\frac{63}{64}\)
\(\Rightarrow G=\frac{5.63}{3.64}=\frac{5.21.3}{3.64}=\frac{5.21}{64}=\frac{105}{64}\)
A = 1 + 2 + 3 + ... + 2018 (có 2018 số )
= (2018 + 1) . 2018 : 2 = 2037171
B = 1 + 3 + 5 + ... + 2017(có 1009 số )
= (2017 + 1) . 1009 : 2 = 1018081
C = 2 + 4 + 6 + ... + 2018 (Có 1009 số )
= (2018 + 2) x 1009 : 2 = 1019090
D = 72 . 153 + 27.153 + 153
= (72 + 27 + 1) . 153
= 100 . 153 = 15300
Đặt A = 1.2 + 2.3 + 3.4 + ... + 98.99
A = 1/3 . ( 1.2.3 + 2.3.3 + 3.4.3 + ... + 98.99.3)
A = 1/3 . [ 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 98.99.(100-97)
A = 1/3 . ( 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 98.99.100 - 97.98.99)
A = 1/3 . [(1.2.3 + 2.3.4 + 3.4.5 + ... + 98.99.100) - ( 0.1.2 + 1.2.3 + 2.3.4 + ... + 97.98.99)]
A = 1/3 . ( 98.99.100 - 0.1.2)
A = 1/3 .98.99.100
A = 323400
Ta có: 323400x/26950 = 12/6/7 : 3/2
12x = 14 × 2/3
12x = 28/3
x = 28/3 : 12
x = 28/3 × 1/12 = 7/9
Vậy x = 7/9
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{15.16}\)
\(\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right).\frac{1}{x}=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{15.16}\right)\)
\(\frac{8+4+2+1}{16}.\frac{1}{x}=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(\frac{15}{16}.\frac{1}{x}=3.\left(1-\frac{1}{16}\right)\)
\(\frac{15}{16}.\frac{1}{x}=3.\frac{15}{16}\)
=> \(\frac{1}{x}=3\)
=> \(x=\frac{1}{3}\)
A=1/1.2+1/2.3+1/3.4+..+1/99.100
=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100
=99/100
A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{9.10}\)
A = \(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+\frac{3}{3}-\frac{3}{4}+...+\frac{3}{9}-\frac{3}{10}\)
A = \(\frac{3}{1}-\frac{3}{10}\)
A = \(\frac{27}{10}\)
Vậy A = \(\frac{27}{10}\)
\(\frac{3}{1\cdot2}+\frac{3}{2\cdot3}+\frac{3}{3\cdot4}+...+\frac{3}{9\cdot10}\)
\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{10}\right)\)
\(=3\frac{9}{10}=\frac{27}{10}\)