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=19/4-37/100+1/8-32/25-5/2+37/12
=(38/8+1/8-20/8)-(37/100+128/100)+37/12
=19/8-155/100+37/12
=469/120
\(4\dfrac{3}{4}+\left(-0,37\right)+\dfrac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\dfrac{1}{12}\)
\(=\dfrac{19}{4}+-\dfrac{37}{100}+\dfrac{1}{8}+-\dfrac{32}{25}+-\dfrac{5}{2}+\dfrac{37}{12}\)
\(=\dfrac{219}{50}+\dfrac{1}{8}+-\dfrac{32}{25}+-\dfrac{5}{2}+\dfrac{37}{12}\)
\(=\dfrac{901}{200}+-\dfrac{32}{25}+-\dfrac{5}{2}+\dfrac{37}{12}\)
\(=\dfrac{129}{40}+-\dfrac{5}{2}+\dfrac{37}{12}\)
\(=\dfrac{29}{40}+\dfrac{37}{12}\)
\(=\dfrac{457}{120}\)
Nhớ ủng hộ 1 Đúng !
a) \(\left(6\dfrac{1}{9}+3\dfrac{7}{11}\right)-1\dfrac{1}{9}\)
\(=\left(\left(6+3\right)+\left(\dfrac{1}{9}+\dfrac{7}{11}\right)\right)-\dfrac{10}{9}\)
\(=\left(9+\dfrac{74}{99}\right)-\dfrac{10}{9}\)
\(=9\dfrac{74}{99}-\dfrac{10}{9}\)
\(=\dfrac{965}{99}-\dfrac{10}{9}\)
\(=\dfrac{95}{11}\)
b) \(1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:5\%\)
\(=1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:\dfrac{1}{20}\)
\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+3\cdot20\)
\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+60\)
\(=\dfrac{1275}{22}\)
c) \(4\dfrac{3}{4}-0,37+\dfrac{1}{8}-1,28-2,5+3\cdot\dfrac{1}{2}\)
\(=\dfrac{19}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{3}{2}\)
\(=\dfrac{89}{40}\)
d) \(\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)
\(=\dfrac{5}{11}+\dfrac{10}{77}-5\cdot\dfrac{2}{11}\)
\(=\dfrac{5}{11}+\dfrac{10}{77}-\dfrac{10}{11}\)
\(=-\dfrac{25}{77}\)
e) \(\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
(câu này chờ mình một chút)
Câu e) anh rút thừa số chung là 2 cùng mũ số ra ngoài thì phân số thành số nguyên. Em nghĩ thế!
\(\left(4\frac{3}{4}+\frac{1}{8}+3\frac{1}{12}\right)-\left(0,37+1,28+2,25\right)\)
\(=7\frac{23}{24}-4,15\)
\(=7\frac{23}{24}-4\frac{3}{20}\)
\(=3\frac{97}{120}\)
Ta có:
\(\text{4/3/4+(-0,37)+1/8+(-1,28)+(-2,5)+3/1/12}\)
\(=4:4:3-0,37+1:8-1,28-2,5+3:1:12\)
\(=\frac{1}{3}-\frac{37}{100}+\frac{1}{8}-\frac{32}{25}-\frac{5}{2}+\frac{1}{4}\)
\(=\frac{200}{600}-\frac{222}{600}+\frac{75}{600}-\frac{768}{600}-\frac{1500}{600}+\frac{150}{600}\)
\(=\frac{200-222+75-768-1500+150}{600}\)
\(=-\frac{2065}{600}\)
\(=-\frac{413}{120}\)
\(4\dfrac{3}{4}+\left(-0,37\right)+\dfrac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\dfrac{1}{12}\)
\(=\left(4\dfrac{3}{4}+\dfrac{1}{8}+3\dfrac{1}{12}\right)+\left[\left(-0,37\right)+\left(-1,28\right)+\left(-2,5\right)\right]\)
\(=\left(\dfrac{19}{4}+\dfrac{1}{8}+\dfrac{37}{12}\right)+\left(-4,15\right)\)
\(=\dfrac{191}{24}-4,15\)
\(=\dfrac{457}{120}=3\dfrac{97}{120}\)
\(4\dfrac{3}{4}+\left(-0,37\right)+\dfrac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\dfrac{1}{12}\)
\(=\dfrac{19}{4}+\left(-\dfrac{37}{100}\right)+\dfrac{1}{8}+\left(-\dfrac{32}{25}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)
=\(\dfrac{475}{100}+\left(-\dfrac{37}{100}\right)+\dfrac{1}{8}+\left(-\dfrac{32}{25}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)
=\(\dfrac{219}{50}+\dfrac{25}{200}+\left(-\dfrac{256}{200}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)
=\(\dfrac{219}{50}+\left(-\dfrac{231}{200}\right)+\left(-\dfrac{30}{12}+\dfrac{37}{12}\right)\)
=