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27 tháng 6 2017

a)\(A=\frac{17}{23}.\frac{8}{16}.\frac{23}{17}.\left(-80\right).\frac{3}{4}\)

    \(A=\left(\frac{17}{23}.\frac{23}{17}\right).\left(\frac{8}{16}.\frac{3}{4}\right).\left(-80\right)\)

     \(A=\frac{3}{8}.\left(-80\right)\)

     \(A=-30\)

b)\(C=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right).\left(\frac{1}{3}+\frac{1}{4}-\frac{7}{12}\right)\)

   \(C=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right).0\)

   \(C=0\)

28 tháng 3 2018

\(A=\frac{17}{23}\cdot\frac{8}{16}\cdot\frac{23}{17}\cdot\left(-80\right)\cdot\frac{3}{4}\)\(=\frac{17\cdot4\cdot2\cdot23\cdot16\cdot\left(-5\right)\cdot3}{23\cdot16\cdot17\cdot4}\)

=> \(A=\frac{2\cdot\left(-5\right)\cdot3}{1}=-30\)

\(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\left(\frac{1}{3}+\frac{1}{4}-\frac{7}{12}\right)\)

=> \(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\left(\frac{7}{12}-\frac{7}{12}\right)\)

=> \(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\cdot0=0\)

11 tháng 4 2019

\(1\frac{13}{15}\cdot3\cdot(0,5)^2\cdot3+\left[\frac{8}{15}-1\frac{19}{60}:1\frac{23}{24}\right]\)

\(=\frac{28}{15}\cdot3\cdot0,5\cdot0,5\cdot3+\left[\frac{8}{15}-\frac{79}{60}:\frac{47}{24}\right]\)

\(=\frac{28}{5}\cdot0,25\cdot3+\left[\frac{32}{60}-\frac{79}{60}\cdot\frac{24}{47}\right]\)

\(=\frac{28}{5}\cdot\frac{25}{100}\cdot3+\left[\frac{32}{60}-\frac{158}{235}\right]\)

\(=\frac{28}{5}\cdot\frac{1}{4}\cdot3+\frac{-98}{705}=\frac{7}{5}\cdot1\cdot3+\frac{-98}{705}\)

Đến đây là tính dễ rồi :v

\((-3,2)\cdot\frac{-15}{64}+\left[0,8-2\frac{4}{15}\right]:1\frac{23}{24}\)

\(=\frac{-32}{10}\cdot\frac{-15}{64}+\left[\frac{8}{10}-\frac{34}{15}\right]:\frac{47}{24}\)

\(=\frac{-32\cdot(-15)}{10\cdot64}+\left[\frac{4}{5}-\frac{34}{15}\right]:\frac{47}{24}\)

\(=\frac{-1\cdot(-3)}{2\cdot2}+\frac{4\cdot3-34}{15}:\frac{47}{24}\)

\(=\frac{3}{4}+\frac{-22}{15}:\frac{47}{24}\)

\(=\frac{3}{4}+\frac{-517}{180}=\frac{-191}{90}\)

Bài 2 : \(\frac{2\cdot(-13)\cdot9\cdot10}{(-3)\cdot4\cdot(-5)\cdot26}=\frac{1\cdot(-1)\cdot3\cdot2}{(-1)\cdot2\cdot(-1)\cdot2}=\frac{1\cdot3}{-1\cdot2}=\frac{3}{-2}=\frac{-3}{2}\)

\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\cdot(8+4)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)

A = (4+\(\frac{1}{5}\)) . \(\frac{18}{19}\)+ (2+\(\frac{8}{5}\)) . \(\frac{21}{5}\)

A= \(\frac{21}{5}\).18/19 + 18/5 . 21/5

A= 21/5 (18/19 + 18/5)

A= 21/5 . 432/95

A= 9288/95

b= 25/2. (3+2/7) - 23/7. (5 + 1/2)

b= 25/2 . 23/7 - 23/7 . 11/2

b= 23/7 (25/2 -11/2)

b=23/7 . 7

b= 23

28 tháng 3 2018

2.  a) \(3^{200}=\left(3^2\right)^{100}=9^{100}\)

          \(2^{300}=\left(2^3\right)^{100}=8^{100}\)

Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)

b) \(71^{50}=\left(71^2\right)^{25}=5041^{25}\)

     \(37^{75}=\left(3^3\right)^{25}=27^{25}\)

Vì \(5041^{25}>27^{25}\Rightarrow71^{50}>37^{75}\)

c) \(\frac{201201}{202202}=\frac{201201:1001}{202202:1001}=\frac{201}{202}\)

      \(\frac{201201201}{202202202}=\frac{201201201:1001001}{202202202:1001001}=\frac{201}{202}\)

Vì \(\frac{201}{202}=\frac{201}{202}\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)

27 tháng 4 2020

Gyvyghghgbhg

30 tháng 7 2015

\(\frac{23.2323.29}{23.292929}=\frac{23.23.101.29}{23.10101.29}=\frac{23.101}{10101}=\frac{2323}{10101}\)

10 tháng 6 2018

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