\(\frac{\frac{4}{115}-\frac{4}{5}-\frac{4}{6115}}{\frac{7}{115}-\frac{7}{5}-\...">
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\(\frac{\frac{4}{115}-\frac{4}{5}-\frac{4}{6115}}{\frac{7}{115}-\frac{7}{5}-\frac{7}{6115}}+\frac{3}{7}\)

\(=\frac{4.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}{7.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}+\frac{3}{7}\)

\(=\frac{4}{7}+\frac{3}{7}=1\)

21 tháng 9 2021

\(a)\)\(\frac{\frac{4}{115}-\frac{4}{5}-\frac{4}{6115}}{\frac{7}{115}-\frac{7}{5}-\frac{7}{6115}}+\frac{3}{7}\)

\(=\)\(\frac{4.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}{7.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}+\frac{3}{7}\)

\(=\)\(\frac{4}{7}+\frac{3}{7}\)

\(=\)\(1\)

17 tháng 12 2016

a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)

= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)

= \(\frac{17}{9}-\frac{2}{3}\)

= \(\frac{11}{9}\)

b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)

= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)

= \(\frac{2}{5}.\frac{7}{12}\)

= \(\frac{7}{30}\)

Mình lười làm quá, hay mình nói kết quả cho bn thôi nha

c) -6

d) 3

e) 3

g) 12

h) \(\frac{23}{18}\)

i) \(\frac{-69}{20}\)

k) \(\frac{-1}{2}\)

l) \(\frac{49}{5}\)

20 tháng 8 2018

b) -5/7 . 4/13 + -5/7 . 9/13 + -2/7

= -5/7 . (4/13 + 9/13) + -2/7

= - 5/7 + - 2/7

= -1

k nếu đúng, học tốt nha

1 tháng 9 2016

a)|-10|:(-2):(-5)+(-3)2

    =1+9

     =10

b)1+(-2)+3+(-4)+5+(-6)+...+21+(-22)

   =[1+(-2)]+[3+(-4)]+[5+(-6)]+...+[21+(-22]

   =(-1)+(-1)+(-1)+...+(-1)

Mà từ 1 đến 22 có:(22-1):1+1:2=11(cặp)

        Suy ra:1+(-2)+3+(-4)+5+(-6)+...+21+(-22)=(-11)

1 tháng 9 2016

c)\(\frac{3}{4}.\frac{5}{9}+\frac{3}{4}.\frac{4}{9}\)

\(=\frac{3}{4}.\left(\frac{5}{9}+\frac{4}{9}\right)\)

\(=\frac{3}{4}\)

d)\(-\frac{4}{17}+\frac{5}{19}+-\frac{13}{17}+\frac{14}{19}+\frac{3}{115}\)

\(=\left[\left(-\frac{4}{17}\right)+\left(-\frac{13}{17}\right)\right]+\left(\frac{5}{19}+\frac{4}{19}\right)+\frac{3}{115}\)

\(=\left(-\frac{27}{17}\right)+1+\frac{3}{115}\)

\(=-\frac{1099}{1955}\)

e)\(\left(\frac{3}{4}+-\frac{7}{2}\right).\left(\frac{10}{11}+\frac{2}{22}\right)\)

\(=\left(\frac{3}{4}-\frac{14}{4}\right).\left(\frac{20}{22}+\frac{2}{22}\right)\)

\(=\left(-\frac{11}{4}\right).\left(\frac{22}{22}\right)\)

\(=-\frac{11}{4}\)

1 tháng 11 2020

a)\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+\frac{1}{2}\)+\(\frac{16}{21}\)

\(=\left(\frac{27}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)\)+\(\frac{1}{2}\)

\(=1+1+\frac{1}{2}\)

\(=2+\frac{1}{2}\)=\(\frac{5}{2}\)

1 tháng 11 2020

DỄ THÌ TỰ LÀM ĐI

18 tháng 10 2018

\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)

\(=\frac{7}{2}-2\)

\(=\frac{7}{2}-\frac{4}{2}\)

\(=\frac{3}{2}\)

\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)

\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)

\(=\frac{3}{7}.\left(2-9\right)\)

\(=\frac{3}{7}.\left(-7\right)\)

\(=-3\)

\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )

18 tháng 10 2018

a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)

\(3\cdot25:\frac{5}{4}\)

\(3\cdot\left(25:\frac{5}{4}\right)\)

=\(3\cdot20\)

=60

b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)

=\(\frac{3}{7}\cdot\left(-7\right)\)

=\(-3\)

c) = 

29 tháng 7 2019

\(a,4\frac{5}{9}:\frac{\left(-5\right)}{7}+\frac{4}{9}:\frac{-5}{7}\)

\(=\frac{41}{9}.\frac{-7}{5}+\frac{4}{9}.\frac{-7}{5}\)

\(=\frac{-7}{5}.\left(\frac{41}{9}+\frac{4}{9}\right)\)

\(=-\frac{7}{9}.5\)

\(=-7\)

a)Bn Kaito Kid làm rùi!

B)Không viết lại đề

\(=\frac{11}{7}\cdot\left(-\frac{3}{5}+\frac{4}{9}-\frac{2}{5}+\frac{5}{9}\right)=\frac{11}{7}\cdot0=0\)

c)Không viết lại đề

\(A=\left(2+4+...+100\right)\left(\frac{3}{5}\cdot\frac{10}{7}-\frac{6}{7}\right):\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(2+4+6+...+100\right)\cdot0\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=0\)

\(=\frac{7}{6}\cdot\left(\frac{3}{26}-\frac{3}{13}+\frac{1}{10}-\frac{8}{5}\right)=\frac{7}{6}\left(\frac{-3}{26}+\frac{-17}{10}\right)=\frac{7}{6}\cdot\frac{236}{130}=\frac{413}{195}\)

D)