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1+1+2+2+3+3+4+4+5+5+6+6+7+7+8+9+8+9+10+12+13+14+15+16+17+18+19+20+30+1000000=10000274
1.3.77−1+3.7.99−3+7.9.1313−7+9.13.1515−9+\frac{19-13}{13.15.19}+13.15.1919−13
=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}=1.31−3.71+3.71−7.91+7.91−9.131+9.131−13.151+13.151−15.191
=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}=1.31−15.191=28595−2851=28594
b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)b,=61.(1.3.76+3.7.96+7.9.136+9.13.156+13.15.196)
làm giống như trên
c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)c,=81.(1.2.31+2.3.41+3.4.51+...+48.49.501)
=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)=161.(1.2.32+2.3.42+3.4.52+...+48.49.502)
=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)=161.(1.2.33−1+2.3.44−2+3.4.55−3+...+48.49.5050−48)
=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)=161.(1.21−2.31+2.31−3.41+3.41−4.51+...+48.491−49.501)
=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}=161.(21−24501)=161.(24501225−24501)=4900153
d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)d,=75.(1.5.87+5.8.127+8.12.157+...+33.36.407)
=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)=75.(1.5.88−1+5.8.1212−5+8.12.1515−8+...+33.36.4040−33)
=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)=75.(1.51−5.81+5.81−8.121+8.121−12.151+...+33.361−36.401)
=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}=75.(51−14401)=75.(1440288−14401)=28841
P/S: . là nhân nha
\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
làm giống như trên
\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)
\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)
P/S: . là nhân nha
\(\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{6}{11}+\frac{16}{11}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)\)
= 1 + 2 + 2
= 5
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
bạn ơi bạn bị sai mất 2 câu cuối rồi đó làm ơn tính lại giùm mình nha
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10=55
2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20=110
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19=100
k mk nha pạn hiền!
10 +1 = 11 có 10 số vậy thì 10 : 2 = 5 thì sẽ bằng 5x11 = 55
2 + 20 = 22 cũng có 10 số lấy 10 : 2 = 5 se bang 22 x5 = 110
1+19 = 20 , 10 : 2 = 5 , 20x5 = 100
a) \(\frac{13}{7}-\frac{1}{2}\times\frac{13}{7}+\frac{3}{2}\times\frac{13}{7}\)
\(=\frac{13}{7}\times\left(1-\frac{1}{2}+\frac{3}{2}\right)\)
\(=\frac{13}{7}\times2\)
\(=\frac{26}{7}\)
b) \(\frac{1}{15}\times\left(\frac{3}{7}+\frac{5}{19}\right)+\frac{3}{7}\times\left(\frac{5}{19}-\frac{1}{15}\right)\)
\(=\frac{1}{15}\times\frac{3}{7}+\frac{1}{15}\times\frac{5}{19}+\frac{3}{7}\times\frac{5}{19}-\frac{3}{7}\times\frac{1}{15}\)
\(=\frac{5}{19}\times\left(\frac{1}{15}+\frac{3}{7}\right)\)
\(=\frac{5}{19}\times\frac{52}{105}\)
\(=\frac{52}{399}\)
c) \(\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}\)
\(=5\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{99\times100}\right)\)
\(=5\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=5\times\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=5\times\frac{49}{100}\)
\(=\frac{49}{20}\)
Lần sau nên đăng ít thôi