\(\frac{\left(\frac{2}{5}+1,15+\frac{-9}{20}\right):0,1}{2\frac{1}{3}+3\frac{1}{6}}\)
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T
22 tháng 2 2020
C = \(25.\left(\frac{-1}{3}\right)^3\) \(+\frac{1}{5}\) \(-2.\left(\frac{-1}{2}\right)^2\) \(-\frac{1}{2}\)
C = \(25.\left(\frac{-1}{27}\right)+\frac{1}{5}\) \(-2.\frac{1}{4}\) \(-\frac{1}{2}\)
C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-\frac{1}{2}\) \(-\frac{1}{2}\)
C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-1\)
C = \(\frac{-125}{135}\) \(+\frac{27}{135}\) \(-\frac{135}{135}\)
C = \(\frac{-233}{135}\)
D = \(-8.\left(\frac{3}{4}-\frac{1}{4}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)
D = \(-8.\frac{1}{2}\) \(.\frac{12}{13}\)
D = \(-4.\frac{12}{13}\)
D = \(\frac{-48}{13}\)
E = \(5\sqrt{16}\) \(-4\sqrt{9}\) \(+\sqrt{25}\) \(-0,3\sqrt{400}\)
E = \(5.4-4.3+5-0,3.20\)
E = \(20-12+5-6\)
E = \(8+\left(-1\right)\)
E = \(7\)
F = \(\left(\frac{-3}{2}\right)\) \(+\left|\frac{-5}{6}\right|\) \(-1\frac{1}{2}\) \(:6\)
F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{3}{2}\) \(.\frac{1}{6}\)
F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{1}{4}\)
F = \(\left(\frac{-18}{12}\right)\) \(+\frac{10}{12}\) \(-\frac{3}{12}\)
F = \(\frac{-11}{12}\)
Chúc cậu hk tốt ~
4 tháng 7 2016
\(\left(\frac{2}{5}+1,15+-\frac{9}{20}\right):0,1\)
\(\left(\frac{2}{5}+\frac{23}{20}+-\frac{9}{20}\right):\frac{1}{10}\)
\(\left(\frac{8}{20}+\frac{23}{20}+-\frac{9}{20}\right)\times\frac{10}{1}\)
\(\frac{22}{20}\times\frac{10}{1}=11\)
\(2\frac{1}{3}+3\frac{1}{6}=\frac{7}{3}+\frac{19}{6}=\frac{14}{6}+\frac{19}{6}=\frac{14+19}{6}=\frac{33}{6}=\frac{11}{2}\)
30 tháng 7 2018
\(a,\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3+1\right)}\)
\(=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.7}=\frac{2.6}{3.7}=\frac{4}{7}\)
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)\(2\frac{17}{20}-1\frac{11}{5}+6\frac{9}{20}:3\)
= \(\frac{57}{20}-\frac{16}{5}+\frac{129}{20}\times\frac13\)
=\(\frac{57}{20}-\frac{16}{5}+\frac{129}{60}\)
=\(\frac{171}{60}-\frac{192}{60}+\frac{129}{60}\)
=\(\frac{108}{60}\)
=\(\frac95\)
\(\frac{\left(\frac{2}{5}+1,15+\frac{-9}{20}\right):0,1}{2\frac{1}{3}+3\frac{1}{6}}=\frac{\left(\frac{2}{5}+\frac{23}{20}+\frac{-9}{20}\right):\frac{1}{10}}{\left(2+3\right)+\left(\frac{1}{3}+\frac{1}{6}\right)}=\frac{\left(\frac{2}{5}+\frac{7}{10}\right):\frac{1}{10}}{5+\frac{1}{2}}\)\(=\frac{\frac{11}{10}:\frac{1}{10}}{\frac{11}{2}}=\frac{11}{\frac{11}{2}}=11:\frac{11}{2}=2\)