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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

\(=\frac{15}{3}=5\)

1 tháng 10 2017

QUÁ DỄ

=5

25 tháng 8 2016

\(\frac{15-\frac{15}{7}-\frac{15}{12}}{3-\frac{3}{7}-\frac{3}{12}}\)

\(=\frac{\frac{90}{7}-\frac{15}{12}}{\frac{18}{7}-\frac{3}{12}}\)

\(=\frac{\frac{325}{28}}{\frac{65}{28}}\)

\(=\frac{35}{17}\)

25 tháng 8 2016

\(\frac{15-\frac{15}{7}-\frac{15}{12}}{3-\frac{3}{7}-\frac{3}{12}}\)

\(=\frac{\frac{90}{7}-\frac{15}{12}}{\frac{18}{7}-\frac{3}{12}}\)

\(=\frac{\frac{325}{28}}{\frac{65}{28}}\)

\(=5\)

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

28 tháng 9 2016

=  15/3

=5

12 tháng 9 2018

\(15-\frac{15}{7}-\frac{15}{12}=11\frac{17}{28}\)

\(=\frac{325}{28}\)

\(3-\frac{3}{7}-\frac{3}{13}=2\frac{31}{91}\)

\(=\frac{213}{91}\)

12 tháng 9 2018

làm từng bước bạn ơi !

15 tháng 9 2017

https://vn.answers.yahoo.com/question/index?qid=20100907070553AAEawEd

lời giải rõ ràng rồi đó bạn ^_~

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 

20 tháng 9 2020

       Bài làm :

Ta có :

\(\frac{15-\frac{15}{7}-\frac{15}{12}}{3-\frac{3}{7}-\frac{3}{12}}\)

\(=\frac{15.\left(1-\frac{1}{7}-\frac{1}{12}\right)}{3.\left(1-\frac{1}{7}-\frac{1}{12}\right)}\)

\(=\frac{15}{3}\)

\(=5\)

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

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