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kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi
\(A=21\frac{4}{11}-\left(1\frac{3}{5}+7\frac{4}{11}\right)\)
\(A=\frac{235}{11}-\left(\frac{8}{5}+\frac{81}{11}\right)\)
\(A=\left(\frac{235}{11}-\frac{81}{11}\right)+\frac{8}{5}\)
\(A=\frac{154}{11}+\frac{8}{5}\)
\(\Rightarrow A=\frac{78}{5}\)
\(B=\left(7\frac{8}{9}+2\frac{3}{13}\right)-\left(4\frac{8}{9}-7\frac{10}{13}\right)\)
\(B=\left(\frac{71}{9}+\frac{29}{13}\right)-\left(\frac{44}{9}-\frac{101}{13}\right)\)
\(B=\left(\frac{71}{9}-\frac{44}{9}\right)+\left(\frac{29}{13}-\frac{101}{13}\right)\)
\(B=\frac{27}{9}+\frac{-72}{13}\)
\(B=3+\frac{-72}{13}\)
\(\Rightarrow B=\frac{-33}{13}\)
P/s: Hoq chắc :v
Bài giải
a, \(\frac{7}{12}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{12}\)
\(=\left(\frac{7}{12}-\frac{5}{12}+\frac{5}{6}+\frac{1}{4}\right)-\frac{3}{7}=\left(\frac{7}{12}-\frac{5}{12}+\frac{10}{12}+\frac{3}{12}\right)-\frac{3}{7}=\frac{5}{4}-\frac{3}{7}=\frac{23}{28}\)
b, \(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\frac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\frac{3^{29}\left(11-3\right)}{3^{28}\cdot4}=\frac{3\cdot8}{4}=6\)
a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)
\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)
\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)
b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)
c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)
e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)
g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)
\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)
A= \(\left(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}\right)+\left(\frac{-6}{13}+\frac{1}{2}+\frac{4}{3}\right)\)
A= \(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}-\frac{6}{13}+\frac{1}{2}+\frac{4}{3}\)
A= \(\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{7}{13}+\frac{6}{13}\right)-\left(\frac{1}{3}-\frac{4}{3}\right)\)
A= \(1-1-\left(-1\right)\)
A= \(1\)
B= \(0,75+\frac{2}{5}+\left(\frac{1}{9}-\frac{7}{5}+\frac{5}{4}\right)\)
B= \(\frac{3}{4}+\frac{2}{5}+\frac{1}{9}-\frac{7}{5}+\frac{5}{4}\)
B= \(\left(\frac{3}{4}+\frac{5}{4}\right)+\left(\frac{2}{5}-\frac{7}{5}\right)+\frac{1}{9}\)
B= \(2-1+\frac{1}{9}\)
B= \(\frac{9}{9}+\frac{1}{9}\)
B= \(\frac{10}{9}\)
C= \(\left(\frac{-3}{2}.\frac{4}{3}\right).\left(\frac{-9}{2}\right)-\frac{1}{4}\)
C = \(-2.\left(\frac{-9}{2}\right)-\frac{1}{4}\)
C = \(9-\frac{1}{4}\)
C = \(\frac{36}{4}-\frac{1}{4}\)
C = \(\frac{35}{4}\)
D = \(\frac{5}{4}.\left(\frac{-7}{10}.\frac{5}{4}-\frac{7}{8}.\frac{7}{10}\right)\)
D = \(\frac{5}{4}.\left(\frac{-7}{8}-\frac{49}{80}\right)\)
D = \(\frac{-35}{32}-\frac{49}{64}\)
D = \(\frac{-70}{64}-\frac{49}{64}\)
D = \(\frac{-119}{64}\)
k mk nha ^_^
a) \(\frac{2}{3}+\frac{1}{3}\cdot\left(-\frac{2}{5}\right)\\ =\frac{2}{3}+\frac{-2}{15}\\ =\frac{10}{15}+\frac{-2}{15}\\ =\frac{8}{15}\)
b) \(0,75\cdot1\frac{7}{9}-1\frac{2}{5}:\frac{-21}{20}\\ =\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}\cdot\frac{-20}{21}\\ =\frac{4}{3}-\frac{-4}{3}\\ =\frac{4}{3}+\frac{4}{3}\\ =\frac{4}{3}\cdot2\\ =\frac{8}{3}\)
c) \(\frac{-2}{17}+\frac{15}{23}+\frac{15}{-17}-\frac{-4}{19}+\frac{8}{23}\\ =\frac{-2}{17}+\frac{15}{23}+\frac{-15}{17}+\frac{4}{19}+\frac{8}{23}\\ =\left(\frac{-2}{17}+\frac{-15}{17}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\\ =\left(-1\right)+1+\frac{4}{19}\\ =0+\frac{4}{19}\\ =\frac{4}{19}\)
d) \(2019^0\cdot\left(6-2\frac{4}{5}\right)\cdot3\frac{1}{8}-1\frac{3}{5}:25\%\\ =1\cdot\left(\frac{30}{5}-\frac{14}{5}\right)\cdot\frac{25}{8}-\frac{8}{5}:\frac{1}{4}\\ =1\cdot\frac{16}{5}\cdot\frac{25}{8}-\frac{8}{5}\cdot4\\ =\frac{16}{5}\cdot\frac{25}{8}-\frac{32}{5}\\ =\frac{50}{5}-\frac{32}{5}\\ =\frac{18}{5}\)
e) \(\left(\frac{7}{8}-\frac{1}{2}\right)\cdot2\frac{2}{3}-\frac{3}{7}\cdot\left(2,5^2\right)\\ =\left(\frac{7}{8}-\frac{4}{8}\right)\cdot\frac{8}{3}-\frac{3}{7}\cdot6,25\\ =\frac{3}{8}\cdot\frac{8}{3}-\frac{3}{7}\cdot\frac{25}{4}\\ =1-\frac{75}{28}\\ =\frac{28}{28}-\frac{75}{28}\\ =\frac{-47}{28}\)
a, \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-2}{5}\right)\)
= \(\frac{2}{3}+\frac{-2}{15}=\frac{8}{15}\)
b, \(0,75.1\frac{7}{9}-1\frac{2}{5}:\frac{-21}{20}\)
= \(\frac{3}{4}.\frac{16}{9}-\frac{7}{5}.\frac{-20}{21}\)
= \(\frac{4}{3}-\left(\frac{-4}{3}\right)=\frac{8}{3}\)
c, \(\frac{-2}{17}+\frac{15}{23}+\frac{15}{-17}+\frac{4}{19}+\frac{8}{23}\)
= \(\left(\frac{-2}{17}+\frac{-15}{17}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\)
= \(\left(-1\right)+1+\frac{4}{19}=0+\frac{4}{19}=\frac{4}{19}\)
d, \(\left(6-2\frac{4}{5}\right).3\frac{1}{8}-1\frac{3}{5}:25\%\)
=> \(\left(6-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}:25\%\)
= \(\frac{16}{5}.\frac{25}{8}-\frac{8}{5}.25:100\)
= 10 - 0,4 = 9,6
e, \(\left(\frac{7}{8}-\frac{1}{2}\right).2\frac{2}{3}-\frac{3}{7}.\left(2,5^2\right)\)
=> \(\frac{3}{8}.\frac{8}{3}-\frac{3}{7}.6,25\)
= \(1-\frac{75}{28}=\frac{-47}{28}\)
a) 4/7 : ( 2/5 . 4/7 )
=4/7 :2/5 : 4/7
=( 4/7 : 4/7 ) : 2/5
=1 . 2/5
=2/5
b)6/7 + 5/7 : 5 -8/9
=6/7 + 1/7 - 8/9
=1 - 8/9
= 1/9
a) \(\frac{4}{7}:\left(\frac{2}{5}\cdot\frac{4}{7}\right)=\frac{4}{7}:\frac{2}{5}:\frac{4}{7}=\frac{4}{7}\cdot\frac{7}{4}\cdot\frac{5}{2}\)\(=1\cdot\frac{5}{2}=\frac{5}{2}\)
b) \(\frac{6}{7}+\frac{5}{7}:5-\frac{8}{9}=\frac{6}{7}+\frac{1}{7}-\frac{8}{9}\)\(=1-\frac{8}{9}=\frac{1}{9}\)