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DT
22 tháng 1 2019
A= 7/8:(4/18-1/18)+7/8:(1/36-15/36)
=7/8:1/6+7/8:(-7/18)
=7/8:(1/6+-7/18)=7/8:(3/18+-7/18)=7/8:(-2/9)=-63/18=-7/2
TM
4 tháng 5 2018
\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)
\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)
\(=\frac{7}{12}+\frac{31}{12}\)
\(=\frac{38}{12}=\frac{19}{6}\)
\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)
\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)
\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\left(\frac{-11}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\frac{2}{13}\)
\(=\frac{-10}{117}\)
\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)
\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)
\(=\frac{-8}{5}-\frac{7}{5}\)
\(=-3\)
\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)
\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)
\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)
\(=\frac{4}{5}\cdot\frac{13}{7}\)
\(=\frac{52}{35}\)
4 tháng 5 2018
a)7/12.6/11+7/12.5/11-2.7/12
=7/12(6/11+5/11-2)
=7/12(1-2)
=7/12.(-1)
=-7/12
7 tháng 8 2019
a . 7/12 . 6/11 + 7/12 . 5/11 - 2 7/12
= 7/12 . ( 6/11 + 5/11 ) - 31/12
= 7/12 . 1 - 31/12
= 7/12 - 31/12
= -2
b . -5/9 . -6/13 + 5/-9 . -5/13 - 5/9
= -5/9 . ( -6/13 + -5/13 ) - 5/9
= -5/9 . ( -1 ) -5/9
= 5/9 - 5/9
= 0
B
2 tháng 4 2019
\(N=\frac{1}{6}.\frac{5}{9}+\frac{5}{9}.\frac{2}{7}+\frac{1}{9}.\frac{5}{7}-\frac{5}{9}.\frac{3}{7}\)
\(N=\frac{1}{6}.\left(\frac{5}{9}+\frac{1}{9}\right).\left(\frac{2}{7}+\frac{5}{7}-\frac{3}{7}\right)\)
\(N=\frac{1}{6}.\frac{6}{9}.\frac{4}{7}\)
\(N=\frac{1}{9}.\frac{4}{7}\)
\(N=\frac{4}{63}\)
12 tháng 6 2020
\(N=\frac{1}{6}.\frac{5}{9}+\frac{5}{9}.\frac{2}{7}+\frac{1}{9}.\frac{5}{7}-\frac{5}{9}.\frac{3}{7}\)
\(N=\frac{1}{6}.\frac{5}{9}+\frac{5}{9}.\frac{2}{7}+\frac{5}{9}.\frac{1}{7}-\frac{5}{9}.\frac{3}{7}\)
\(N=\frac{5}{9}.\left(\frac{1}{6}+\frac{2}{7}+\frac{1}{7}-\frac{3}{7}\right)\)
\(N=\frac{5}{9}.\frac{1}{6}\)
\(N=\frac{5}{54}\)
11 tháng 8 2019
Đặt P = ... ( biểu thức đề bài )
Nhận xét: Với \(k\inℕ^∗\) ta có:
\(\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}=\frac{k+2}{k!+\left(k+1\right).k!+\left(k+2\right).k!}=\frac{k+2}{2.k!\left(k+2\right)}=\frac{1}{2.k!}\)
\(\Rightarrow\)\(P=\frac{1}{2.1!}+\frac{1}{2.2!}+...+\frac{1}{2.6!}=\frac{1}{2}\left(1+\frac{1}{2}+...+\frac{1}{720}\right)=...\)
\(A=\frac{6}{7}+\frac{1}{7}.\frac{2}{7}+\frac{1}{7}.\frac{5}{7}\)
\(A=\frac{6}{7}+\frac{1}{7}.\left(\frac{2}{7}+\frac{5}{7}\right)\)
\(A=\frac{6}{7}+\frac{1}{7}.1\)
\(A=\frac{6}{7}+\frac{1}{7}\)
\(A=\frac{7}{7}=1\)
\(A=\frac{6}{7}+\frac{1}{7}.\frac{2}{7}+\frac{1}{7}.\frac{5}{7}=\frac{1}{7}.\left(6+\frac{2}{7}+\frac{5}{7}\right)=\frac{1}{7}.7=1\)