NT
Nguyễn Thị Thương Hoài
Giáo viên
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21 tháng 3
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D
12 tháng 5 2015
Đặt A = \(\frac{\frac{1}{2}}{1+2}+\frac{\frac{1}{2}}{1+2+3}+...+\frac{\frac{1}{2}}{1+2+3+....+100}\)
= \(\frac{1}{2}\left(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{100.101:2}\right)\)
= \(\frac{1}{2}\left(\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{100.101}\right)\)
= \(\frac{1}{2}.2\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\right)\)
= 1\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{100}-\frac{1}{101}\right)\)
= \(\frac{1}{2}-\frac{1}{101}=\frac{101}{202}-\frac{2}{202}=\frac{99}{202}\)
19 tháng 7 2016
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}+\frac{1}{3}\right)=\frac{4}{5}.\frac{3}{3}=\frac{4}{5}.1=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}+\frac{1}{6}\right)=\frac{3}{4}:\frac{2}{3}=\frac{9}{8}\)
\(\frac{2}{3}.\frac{4}{5}-\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}-\frac{1}{3}\right)=\frac{4}{5}.\frac{1}{3}=\frac{4}{15}\)
\(\frac{1}{2}:\frac{3}{4}-\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{3}{4}:\frac{1}{3}=\frac{9}{4}\)
19 tháng 7 2016
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\left(\frac{2}{3}+\frac{1}{3}\right).\frac{4}{5}=1.\frac{4}{5}=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{1}{2}.\frac{4}{3}+\frac{1}{6}.\frac{4}{3}=\left(\frac{1}{2}+\frac{1}{6}\right).\frac{4}{3}=\frac{2}{3}.\frac{4}{3}=\frac{8}{9}\)
c,d tương tự
7 tháng 11 2017
\(a.\) \(3\frac{3}{4}+\left(4\frac{2}{4}-3\frac{1}{2}\right):\frac{3}{4}\)
\(=\frac{15}{4}\left(\frac{18}{4}-\frac{7}{2}\right):\frac{3}{4}\)
\(=\frac{15}{4}+\frac{4}{4}:\frac{3}{4}=\frac{15}{4}+\frac{4}{3}\)
\(=\frac{15}{4}+\frac{4}{3}=\frac{61}{12}\)
\(b.\) \(7\cdot\frac{2}{3}-\frac{2}{5}:\frac{1}{2}-\frac{2}{3}\)
\(=\frac{7}{3}-\frac{2}{5}\cdot2-\frac{2}{3}\)
\(=\frac{7}{3}-\frac{4}{5}-\frac{2}{3}=\frac{35-12-10}{15}\)
\(=\frac{13}{15}\)
\(c.\) \(68,7-100:20+70,8\)
\(=68,7-5+70,8\)
\(=63,7+70,8\)
\(=134,5\)
\(d.\)\(\left(5915+445:5\right)-76\cdot25\)
\(=\left(5915+89\right)-1900\)
\(=6004-1900=4104\)
7 tháng 11 2017
a)=\(\frac{15}{4}+\left(\frac{18}{4}-\frac{7}{2}\right)\)/ \(\frac{3}{4}\)
=\(\left(\frac{33}{4}-\frac{14}{4}\right)\)/ \(\frac{3}{4}\)
= \(\frac{19}{4}\cdot\frac{4}{3}\)
=\(\frac{19}{3}\)
b) = \(\frac{2}{3}\cdot\left(7-1\right)-\frac{2}{5}\cdot\frac{2}{1}\)
= \(\frac{2}{3}\cdot6-\frac{4}{5}\)
= 1-\(\frac{4}{5}\)
= \(\frac{1}{5}\)
c) = 68.7 - 5 + 70.8
= 63.7 + 70.8
=134.5
d) = (5915 - 89) -76*25
= 5826 - 1900
= 3926
MA
14 tháng 10 2016
a ) \(1\frac{1}{2}+2\frac{1}{3}+3\frac{1}{6}-5\)
\(=\frac{3}{2}+\frac{7}{3}+\frac{19}{6}-\frac{5}{1}\)
\(=\frac{9}{6}+\frac{14}{6}+\frac{19}{6}-\frac{30}{6}\)
\(=\frac{23}{6}+\frac{19}{6}-\frac{30}{6}\)
\(=\frac{42}{6}-\frac{30}{6}\)
\(=\frac{12}{6}=2\)
b ) \(2\frac{2}{3}\times3\frac{3}{4}\div4\frac{4}{5}\)
\(=\frac{8}{3}\times\frac{15}{4}\div\frac{24}{5}\)
\(=\frac{120}{12}\div\frac{24}{5}\)
\(=\frac{120}{12}\times\frac{5}{24}\)
\(=\frac{600}{288}=\frac{25}{12}\)
c ) \(4\frac{1}{5}+5\frac{1}{3}-2\frac{2}{3}\times3\frac{1}{5}+\frac{9}{25}\div\frac{9}{20}\)
\(=\frac{21}{5}+\frac{16}{3}-\frac{8}{3}\times\frac{16}{5}+\frac{9}{25}\div\frac{9}{20}\)
\(=\frac{63}{15}+\frac{80}{15}-\frac{128}{15}+\frac{9}{25}\times\frac{20}{9}\)
\(=\frac{143}{15}-\frac{128}{15}+\frac{180}{225}\)
\(=\frac{15}{15}+\frac{12}{15}\)
\(=\frac{27}{15}=\frac{9}{5}\)
