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\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
1: =7/2+23/5=35/10+46/10=81/10
2: =6/8+53/8=59/8
3: =43/8+65/7=821/56
a, \(\dfrac{x}{2}+\dfrac{3x}{4}=\dfrac{4}{5}\Leftrightarrow\dfrac{10x+15x}{20}=\dfrac{16}{20}\Rightarrow25x=16\Leftrightarrow x=\dfrac{16}{25}\)
b, \(\dfrac{3}{7}.\dfrac{5}{8}-\dfrac{3}{8}.\dfrac{13}{8}+\dfrac{1}{7}=\dfrac{15}{56}-\dfrac{39}{64}+\dfrac{1}{7}\)
\(=\dfrac{120}{448}-\dfrac{273}{448}+\dfrac{64}{448}=-\dfrac{89}{448}\)
a) 5 + (-8) . 3 = 5 + (-24) = -19
b) 4 + - 5 2 = 4 + 25 = 29
c) 1 – 2 – 3 + 4 + 5 – 6 – 7 + 8 + … + 801 – 802 – 803 + 804
= (1 – 2 – 3 + 4) + (5 – 6 – 7 + 8) + … + (801 – 802 – 803 + 804)
= 0 + 0 + … + 0 = 0
1-2+3-4+5-6+...+51-52+53
=(1-2)+(3-4)+...+(51-52)+53
=(-1)+(-1)+...+(-1)+53
=(-1)×26+53
=-26+53
=27
1-2+3-4+5-6+...+51-52+53
=(1-2)+(3-4)+...+(51-52)+53
=(-1)+(-1)+...+(-1)+53
=(-1)×26+53
=-26+53
=27
Sửa đề: \(\dfrac{\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
=1
a: =>5x+25-3x+6=25+18
=>2x+41=43
=>2x=2
=>x=1
b: =>4x+8=3x+3+17
=>4x+8=3x+20
=>x=12
a: =>5x+25-3x+6=25+18
=>2x+41=43
=>2x=2
=>x=1
b: =>4x+8=3x+3+17
=>4x+8=3x+20
=>x=12
S = 1 x 3 + 2 x 4 + 3 x 5 + 4 x6 + ...+ 49 x 51 + 50 x 52
S = ( 1 x3 + 3 x5 + ..+ 49x51) + (2x4+4x6+...+50x52)
Đặt A = 1x3+3x5+...+49x51
=> 6A = 1x3x6+3x5x6+...+49x51x6
6A = 1x3x(5+1) + 3x5x(7-1) + ...+ 49x51x(53-47)
6A = 1x3x5 + 1x3 + 3x5x7 - 1x3x5 + ...+ 49x51x53 - 47x49x51
6A = (1x3 + 1x3x5 + 3x5x7+...+49x51x53) - (1x3x5+...+47x49x51)
6A = 1x3 + 49x51x53
A = 22 075
Tương tự như trên ta có: B = 2x4 + 4x6 + ...+ 50x52
B = 23 400
Thay B ;A vào S
S = 22 075 +23 400
S = 45 475
2=10
3=15
4=20
5=25
2=10,3=15,4=20,=5=1