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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) 3 x X x 2/3 = 4/5
3.x=6/5
x=2/5
b) 3,94 x 18,24 + 18,24 x 3,72 + 18,24 x 2,34
=18,24(3,94+3,72+2,34)
=18,24.10
=182,4
d) 8/19 x 2/7 + 8/19 x 1/7 + 8/19 x 4/7
=8/19(2/7+1/7+4/7)
=8/19.1
=8/19
8
vì 3+1+7=11
5+3+9=17
số trung bình cộng là: (11+17):2=14
4+2+?=14
?=14-(4+2)
?=14-6
?=8
mình làm thế có đúng ko các bạn
1+2+2+3+3+4+4+5+5+6+6+6+7+7+8+8+8+0+9+9+90+70+496+454+6533+7545x1000x34343-65464+753256x123= 2,592105278x1011
1+2+2+3+3+4+4+5+5+6+6+6+7+7+8+8+8+0+9+9+90+70+496+454+6533+7545x1000x34343-65464+753256x123= 2,592105278x1011
a)\(\frac{4}{5}-\frac{1}{4}+\frac{3}{10}\)
\(=\frac{16}{20}-\frac{5}{20}+\frac{6}{20}\)
\(=\frac{17}{20}\)
b) \(\frac{2}{5}:\left(1-\frac{1}{10}\right)\)
\(=\frac{2}{5}:\frac{9}{10}\)
\(=\frac{4}{9}\)
c)\(\frac{7}{8}\times\frac{4}{9}+\frac{1}{14}:\frac{5}{14}\)
\(=\frac{7}{18}+\frac{1}{5}\)
\(=\frac{53}{90}\)
d)\(\frac{2}{7}\times\frac{3}{11}+\frac{2}{7}\times\frac{8}{11}\)
\(=\frac{2}{7}\times\left(\frac{3}{11}+\frac{8}{11}\right)\)
\(=\frac{2}{7}\times1=\frac{2}{7}\)
e) \(12+\left(16-11\right)\times4\)
\(=12+20=32\)
f)\(2\frac{3}{7}+1\frac{4}{7}\)
\(=\frac{17}{7}+\frac{11}{7}\)
\(=4\)
g)\(\frac{2}{3}\times\frac{4}{5}+\frac{1}{5}:\frac{9}{11}\)
\(=\frac{8}{15}+\frac{11}{45}\)
\(=\frac{7}{9}\)
h)\(\left(6,2:2+3,7\right):0,2\)
\(=\left(3,1+3,7\right):0,2\)
\(=6,8:0,2=34\)
#H
a) \(\frac{1.3+3.5+5.7+7.9}{3.6+9.10+15.14+21.18}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.2.3+3.5.2.3+5.7.2.3+7.9.2.3}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.6+3.5.6+5.7.6+7.9.6}\)
= \(\frac{1.3+3.5+5.7+7.9}{6.\left(1.3+3.5+5.7+7.9\right)}=\frac{1}{6}\)
Dấu "." là dấu nhân cấp 2
b) \(\frac{1.2+2.3+3.4+4.5}{3.6+6.9+9.12+12.15}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.3.3+2.3.3.3+3.4.3.3+4.5.3.3}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.9+2.3.9+3.4.9+4.5.9}\)
= \(\frac{1.2+2.3+3.4+4.5}{9.\left(1.2+2.3+3.4+4.5\right)}=\frac{1}{9}\)
Dấu "." là dấu nhân cấp 2
c) \(\frac{0,3+\frac{3}{7}+\frac{3}{11}}{0,4+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{\frac{3}{10}+\frac{3}{7}+\frac{3}{11}}{\frac{4}{10}+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{3.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}{4.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}=\frac{3}{4}\)
\(12\frac{14}{25}\cdot6\frac{10}{12}=12\frac{14}{25}\cdot6\frac{5}{6}=\frac{314}{25}\cdot\frac{41}{6}=\frac{314\cdot41}{25\cdot6}=\frac{157\cdot2\cdot41}{25\cdot3\cdot2}=\frac{6437}{75}\)
\(5\frac{6}{9}\div7\frac{6}{3}=ERROR\)
Ta có:
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(A=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{10}\right)=2.\frac{9}{10}\)
\(B=\frac{9}{5}\)
1 + 4 = 5
2 + 7 = 14
4 + 9 = 27
5 + 7 = 39
đố cái quái chi rứa hỉ 5 +7 = 40