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VT
4
HT
6 tháng 9 2015
\(\frac{15-\frac{15}{7}-\frac{15}{12}}{3-\frac{3}{7}-\frac{3}{12}}\)
= \(\frac{15.\left(1-\frac{1}{7}-\frac{1}{12}\right)}{3.\left(1-\frac{1}{7}-\frac{1}{12}\right)}\)
= \(\frac{15}{3}\)
= 5
O
2
15 tháng 9 2017
https://vn.answers.yahoo.com/question/index?qid=20100907070553AAEawEd
lời giải rõ ràng rồi đó bạn ^_~
HT
15 tháng 9 2017
\(\frac{15-\frac{15}{7}-\frac{15}{12}}{3-\frac{3}{7}-\frac{3}{12}}\)
\(=\frac{15\left(1-\frac{1}{7}-\frac{1}{12}\right)}{3\left(1-\frac{1}{7}-\frac{1}{12}\right)}\)
\(=\frac{15}{3}\)
K nha
VB
2
TA
1 tháng 10 2017
\(\frac{15-\frac{15}{7}-\frac{15}{2}}{3-\frac{3}{7}-\frac{3}{12}}\)
\(=\frac{15.\left(1-\frac{1}{7}-\frac{1}{2}\right)}{3.\left(1-\frac{1}{7}-\frac{1}{2}\right)}\)mình sửa đề luôn nhé
\(=\frac{15}{3}=5\)
TP
3
B
20 tháng 9 2020
Bài làm :
Ta có :
\(\frac{15-\frac{15}{7}-\frac{15}{12}}{3-\frac{3}{7}-\frac{3}{12}}\)
\(=\frac{15.\left(1-\frac{1}{7}-\frac{1}{12}\right)}{3.\left(1-\frac{1}{7}-\frac{1}{12}\right)}\)
\(=\frac{15}{3}\)
\(=5\)
20 tháng 9 2020
Bài làm :
\(\frac{15-\frac{15}{7}-\frac{15}{12}}{3-\frac{3}{7}-\frac{3}{12}}\)
\(=\frac{15.\left(1-\frac{1}{7}-\frac{1}{12}\right)}{3.\left(1-\frac{1}{7}-\frac{1}{12}\right)}\)
\(=\frac{15}{3}=5\)
Học tốt
H
0
12 tháng 9 2018
\(15-\frac{15}{7}-\frac{15}{12}=11\frac{17}{28}\)
\(=\frac{325}{28}\)
\(3-\frac{3}{7}-\frac{3}{13}=2\frac{31}{91}\)
\(=\frac{213}{91}\)
FT
0
10 tháng 8 2022
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
\(\frac{15-\frac{15}{7}-\frac{15}{12}}{3-\frac{3}{7}-\frac{3}{12}}\)
\(=\frac{\frac{90}{7}-\frac{15}{12}}{\frac{18}{7}-\frac{3}{12}}\)
\(=\frac{\frac{325}{28}}{\frac{65}{28}}\)
\(=\frac{35}{17}\)
\(\frac{15-\frac{15}{7}-\frac{15}{12}}{3-\frac{3}{7}-\frac{3}{12}}\)
\(=\frac{\frac{90}{7}-\frac{15}{12}}{\frac{18}{7}-\frac{3}{12}}\)
\(=\frac{\frac{325}{28}}{\frac{65}{28}}\)
\(=5\)