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DP
18 tháng 6 2017
a, \(A=\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)
\(A=\left(\frac{2}{5}-\frac{3}{4}\right)+\left(\frac{-1}{6}-\frac{-2}{3}\right)\)
\(A=\left(\frac{8}{20}-\frac{15}{20}\right)+\left(\frac{-3}{18}-\frac{-12}{18}\right)\)
\(A=\frac{-7}{20}+\frac{1}{2}\)
\(\Rightarrow A=\frac{-7}{20}+\frac{10}{20}=\frac{3}{20}\)
DP
18 tháng 6 2017
b, \(B=\frac{7}{10}-\frac{-3}{4}+\frac{-5}{6}-\frac{1}{5}+\frac{-2}{3}\)
\(B=\left(\frac{7}{10}-\frac{1}{5}\right)+\left(\frac{-5}{6}+\frac{-2}{3}\right)-\frac{-3}{4}\)
\(B=\left(\frac{7}{10}-\frac{2}{10}\right)+\left(\frac{-5}{6}+\frac{-4}{6}\right)-\frac{-3}{4}\)
\(B=\frac{1}{2}+\frac{-3}{2}-\frac{-3}{4}\)
\(B=\frac{2}{4}+\frac{-6}{4}-\frac{-3}{4}\)
\(\Rightarrow B=\frac{2+-6+3}{4}=\frac{-1}{4}\)
c, \(C=\frac{\left(\frac{1}{2}-0,75\right)\times\left(0,2-\frac{2}{5}\right)}{\frac{5}{9}-1\frac{1}{12}}\)
\(C=\frac{\left(\frac{1}{2}-\frac{3}{4}\right)\times\left(\frac{1}{5}-\frac{2}{5}\right)}{\frac{5}{9}-\frac{1\times12+1}{12}}\)
\(C=\frac{\left(\frac{2}{4}-\frac{3}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{5}{9}-\frac{13}{12}}\)
\(C=\frac{\left(\frac{-1}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{60}{108}-\frac{117}{108}}\)
\(C=\frac{\frac{1}{20}}{\frac{-19}{36}}=\frac{1}{20}\div\frac{-19}{36}=\frac{1}{20}\times\frac{36}{-19}\)
\(\Rightarrow C=\frac{36}{-380}=\frac{-9}{95}\)
d, \(D=\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{4}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(D=\frac{\frac{56}{84}+\frac{24}{84}-\frac{21}{84}}{\frac{-10}{7}+\frac{3}{28}}\)
\(D=\frac{\frac{59}{84}}{\frac{-40}{28}+\frac{2}{28}}=\frac{59}{84}\div\frac{-37}{28}=\frac{59}{84}\times\frac{28}{-37}\)
\(\Rightarrow D=\frac{1652}{-3108}=\frac{-59}{111}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 3
B1a)\(11\frac34-\left(6\frac56-4\frac12\right)+1\frac23\)
=\(11\frac34-6\frac56+4\frac12+1\frac23\)
=\(\left(11-6+4+1\right)+\left(\frac34-\frac56+\frac12+\frac23\right)\)
=\(10+\left(\frac{9}{12}-\frac{10}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=\(10+\left(-\frac{1}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=10+\(\frac{13}{12}\)
=\(\frac{120}{12}+\frac{13}{12}\)
=\(\frac{133}{12}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 3
b)\(2\frac{17}{20}-1\frac{11}{5}+6\frac{9}{20}:3\)
= \(\frac{57}{20}-\frac{16}{5}+\frac{129}{20}\times\frac13\)
=\(\frac{57}{20}-\frac{16}{5}+\frac{129}{60}\)
=\(\frac{171}{60}-\frac{192}{60}+\frac{129}{60}\)
=\(\frac{108}{60}\)
=\(\frac95\)
12 tháng 7 2017
a)\(\frac{2}{3}+\frac{3}{4}+\frac{5}{6}\)
\(=\frac{8+9+10}{12}\)
\(=\frac{27}{12}=\frac{9}{4}\)
b)\(\frac{15}{8}-\frac{7}{12}+\frac{5}{6}\)
\(=\frac{45-14+20}{24}\)
\(=\frac{51}{24}=\frac{17}{8}\)
2)
a)\(\frac{2}{5}+\frac{7}{13}+\frac{3}{5}+\frac{1}{7}\)
\(=\frac{2}{5}+\frac{3}{5}+\frac{7}{13}+\frac{1}{7}\)
\(=1+\frac{7}{13}+\frac{1}{7}\)
\(=\frac{20}{13}+\frac{1}{7}\)
\(=\frac{153}{91}\)
Tí tớ trả lời tiếp
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
5 tháng 3
b)\(5\frac14+3\frac25-4\frac14\)
=\(\left(5\frac14-4\frac14\right)+3\frac25\)
=\(\left\lbrack\left(5-4\right)+\left(\frac14-\frac14\right)\right\rbrack+\frac{17}{5}\)
=\(1+0+\frac{17}{5}\)
=\(\frac55+\frac{17}{5}\)
=\(\frac{22}{5}\)
PT
26 tháng 4 2017
kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi
HH
20 tháng 8 2020
\(a)\frac{5}{6}+\left(\frac{-1}{2}\right)+\frac{3}{4}\)
\(=\frac{1}{3}+\frac{3}{4}\)
\(=\frac{13}{12}\)
\(b)\left(0,75-\frac{1}{3}\right):\frac{7}{15}\)
\(=\left(\frac{3}{4}-\frac{1}{3}\right).\frac{15}{7}\)
\(=\frac{5}{12}.\frac{15}{7}\)
\(=\frac{25}{28}\)
\(c)\frac{7}{12}-\frac{3}{4}.\frac{5}{6}\)
\(=\frac{7}{12}-\frac{5}{8}\)
\(=\frac{-1}{24}\)
\(d)\left(2\frac{1}{3}+1\frac{3}{4}\right).\frac{12}{13}\)
\(=\left(\frac{7}{3}+\frac{7}{4}\right).\frac{12}{13}\)
\(=\frac{49}{12}.\frac{12}{13}\)
\(=\frac{49}{13}\)
20 tháng 8 2020
a)\(\frac{5}{6}+\frac{-1}{2}+\frac{3}{4}=\frac{10}{12}-\frac{6}{12}+\frac{9}{12}=\frac{10-6+9}{12}=\frac{13}{12}\)
b)\(\left\{0,75-\frac{1}{3}\right\}:\frac{7}{15}=\left\{\frac{3}{4}-\frac{1}{3}\right\}.\frac{15}{7}=\left\{\frac{9}{12}-\frac{4}{12}\right\}.\frac{15}{7}=\frac{5}{12}.\frac{15}{7}=\frac{75}{84}\)
c)\(\frac{7}{12}-\frac{3}{4}.\frac{5}{6}=\frac{7}{12}-\frac{5}{8}=\frac{14}{24}-\frac{15}{24}=\frac{-1}{24}\)
d)\(\left\{2\frac{1}{3}+1\frac{3}{4}\right\}.\frac{12}{13}=\left\{\frac{7}{3}+\frac{7}{4}\right\}.\frac{12}{13}=\left\{\frac{28}{12}+\frac{21}{12}\right\}.\frac{12}{13}=\frac{39}{12}.\frac{12}{13}=3\)
HJ
7 tháng 8 2017
\(a,\left(10\frac{2}{9}.2\frac{3}{5}\right)-6\frac{2}{9}=\frac{1196}{45}-\frac{56}{9}=\frac{1196}{45}-\frac{280}{45}=\frac{916}{45}\)
\(b,\frac{6}{7}+\frac{1}{7}.\frac{2}{7}+\frac{1}{7}.\frac{5}{7}=\frac{1}{7}\left(6+\frac{2}{7}+\frac{5}{7}\right)=\frac{1}{7}.7=1\)
\(c,3.136.8+4.14.6-14.150=3264+336-2100=1500\)
\(d,\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
\(e,\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)
7 tháng 7 2017
-\(\frac{-2}{3}+\frac{3}{4}-\frac{-1}{6}+\frac{-2}{5}=-\frac{4}{6}+\frac{1}{6}+\frac{3}{4}-\frac{2}{5}=-\frac{2}{4}+\frac{3}{4}-\frac{2}{5}=\frac{1}{4}-\frac{2}{5}=-\frac{3}{20}\)
= \(-\frac{3}{20}\)
B
24 tháng 3 2019
a) \(\frac{14}{21}+1-\left|\frac{1}{3}-1\right|\)
\(=\frac{2}{3}+1-\frac{2}{3}\)
\(=1+\left(\frac{2}{3}-\frac{1}{3}\right)\)
\(=1\)
b) \(\frac{1}{3}-\left|\frac{-1}{4}+\frac{5}{6}\right|-\left|\frac{-7}{12}\right|\)
\(=\frac{1}{3}-\frac{7}{12}-\frac{7}{12}\)
\(=-\frac{5}{6}\)
A = \(\frac{1}{2}+\frac{-2}{5}=\frac{1}{10}\)
B = \(\frac{2}{7}-\frac{6}{7}:3-3:\frac{2}{3}=-\frac{9}{2}\)
C = Câu hỏi này không thể tính được