\(l.\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{2}{3}+\dfra...">
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6 tháng 6 2023

\(\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{23}}\times\dfrac{\dfrac{1}{3}-0,25-0,2}{1\dfrac{1}{6}-0,875-0,7}\)

\(=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{23}-\dfrac{1}{23}}\times\dfrac{\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}-\dfrac{7}{10}}\)

\(=\dfrac{\dfrac{1}{3} +\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{1}{3}\times2+\dfrac{1}{7}\times2-\dfrac{1}{23}\times2}\times\dfrac{\dfrac{2}{6}-\dfrac{2}{8}-\dfrac{2}{10}}{7\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}\)

\(=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{2\times\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}\right)}\times\dfrac{2\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}{7\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}\)

\(=\dfrac{1}{2}\times\dfrac{2}{7}\)

\(=\dfrac{1}{7}\)

6 tháng 6 2023

giỏi waaaaaa

Bài 1: 

a: \(A=\left(-\dfrac{1}{5}\right)^{33}:\left(-\dfrac{1}{5}\right)^{32}=\dfrac{-1}{5}\)

c: \(C=\dfrac{2^{12}\cdot3^{10}+3^9\cdot2^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)

a: \(=5-2\cdot\dfrac{1}{4}=5-\dfrac{1}{2}=\dfrac{9}{2}\)

b: \(=\left(\dfrac{7}{2}\right)^3+\dfrac{1}{2}=\dfrac{343}{8}+\dfrac{1}{2}=\dfrac{347}{8}\)

c: \(=\left(5+\dfrac{5}{27}-\dfrac{5}{27}\right)+\left(\dfrac{7}{23}+\dfrac{16}{23}\right)-\dfrac{1}{2}=5+1-\dfrac{1}{2}=5+\dfrac{1}{2}=5.5\)

e: \(=\dfrac{-5}{4}\left(35+\dfrac{1}{6}-45-\dfrac{1}{6}\right)=\dfrac{-5}{4}\cdot\left(-10\right)=\dfrac{50}{4}=\dfrac{25}{2}\)

16 tháng 10 2017

Lời giải :

a ) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)

\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)

\(=2,5\)

b ) \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)

\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)

\(=\dfrac{3}{7}\left(19-33\right)\)

\(=\dfrac{3}{7}\left(-14\right)\)

\(=-6\)

c ) \(9\left(-\dfrac{1}{3}\right)^3+\dfrac{1}{3}\)

\(=9\left(-\dfrac{1}{27}\right)+\dfrac{1}{3}\)

\(=-\dfrac{1}{3}+\dfrac{1}{3}\)

\(=0\)

d ) \(15\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}\div\left(-\dfrac{5}{7}\right)\)

\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right)\div\left(-\dfrac{5}{7}\right)\)

\(=-10\left(-\dfrac{7}{5}\right)\)

\(=14\)

25 tháng 12 2017

Tính 1 câu thoy nhé !

\(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)

= \(\dfrac{3}{7}.\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)

=\(\dfrac{3}{7}.-14=-6\)

21 tháng 11 2017

lấy casio mà tính cho nhanhbanh

1 tháng 12 2017

1 like

26 tháng 11 2022

a: \(=\dfrac{-3}{4}\left(31+\dfrac{11}{23}+8+\dfrac{12}{23}\right)=\dfrac{-3}{4}\cdot40=-30\)

b: \(=\left(\dfrac{7}{3}+\dfrac{7}{2}\right):\left(-\dfrac{25}{6}+\dfrac{22}{7}\right)+\dfrac{15}{2}\)

\(=\dfrac{35}{6}:\dfrac{-175+132}{42}+\dfrac{15}{2}\)

\(=\dfrac{35}{6}\cdot\dfrac{42}{-43}+\dfrac{15}{2}\)

\(=\dfrac{35\cdot7}{-43}+\dfrac{15}{2}\)

\(=\dfrac{-70\cdot7+15\cdot43}{86}=\dfrac{155}{86}\)

c: \(=\dfrac{-7}{5}\left(4+\dfrac{5}{9}+5+\dfrac{4}{9}\right)=\dfrac{-7}{5}\cdot10=-14\)

d: \(=4+\dfrac{25}{16}+25\cdot\left(\dfrac{9}{16}\cdot\dfrac{64}{125}\cdot\dfrac{-8}{27}\right)\)

\(=\dfrac{89}{16}+25\cdot\dfrac{-32}{375}\)

\(=\dfrac{89}{16}-\dfrac{32}{15}=\dfrac{823}{240}\)

e: \(=\dfrac{2}{3}-4\cdot\left(\dfrac{2}{4}+\dfrac{3}{4}\right)=\dfrac{2}{3}-5=-\dfrac{13}{3}\)

\(=5+\dfrac{1}{5}-\dfrac{2}{9}-2+\dfrac{1}{23}+\dfrac{73}{35}-\dfrac{5}{6}-8-\dfrac{2}{7}+\dfrac{1}{18}\)

\(=-5+\left(\dfrac{1}{5}-\dfrac{2}{7}+\dfrac{73}{35}\right)+\left(\dfrac{-2}{9}-\dfrac{5}{6}+\dfrac{1}{18}\right)+\dfrac{1}{23}\)

\(=-5+\dfrac{7-10+73}{35}+\dfrac{-4-15+1}{18}+\dfrac{1}{23}\)

\(=-5+2-1+\dfrac{1}{23}=-4+\dfrac{1}{23}=-\dfrac{91}{23}\)

13 tháng 7 2018

\(a)\dfrac{-5}{21}-\dfrac{1}{3}+3\dfrac{1}{2}.\left(\dfrac{-2}{3}\right)^3\)

\(=\dfrac{-5}{21}+\dfrac{-7}{21}+\dfrac{7}{2}.\dfrac{-8}{27}\)

\(=-\dfrac{4}{7}+\dfrac{-28}{27}\)

\(=\dfrac{-108}{189}+\dfrac{-196}{189}\)

\(=-\dfrac{304}{189}\)

14 tháng 7 2018

\(b)-2\dfrac{1}{3}+\left(\dfrac{3}{8}-\dfrac{3}{4}\right)^3:\dfrac{5}{9}-\dfrac{1}{2}\)

\(=-\dfrac{7}{3}+\left(\dfrac{3}{8}-\dfrac{6}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)

\(=-\dfrac{7}{3}+\left(-\dfrac{3}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)

\(=-\dfrac{7}{3}+\dfrac{-27}{512}.\dfrac{9}{5}-\dfrac{1}{2}\)

\(=-\dfrac{7}{3}+\dfrac{-243}{2560}-\dfrac{1}{2}\)

\(=\dfrac{-17920}{7680}+\dfrac{-729}{7680}+\dfrac{-3840}{7680}\)

\(=\dfrac{-22489}{7680}\)

6 tháng 12 2017

a. \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)

\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)

\(=1+1+0,5\)

\(=2,5\)

b. \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)

\(=\dfrac{3}{7}.\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)

\(=\dfrac{3}{7}.\left(-14\right)=-6\)

c. \(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(\dfrac{-5}{7}\right)\)

\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)\)

\(=-10:\left(-\dfrac{5}{7}\right)\)

\(=14\)

d. \(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\dfrac{-5}{21}:\dfrac{4}{5}+\dfrac{5}{21}:\dfrac{4}{5}\)

\(=\left(\dfrac{-5}{7}+\dfrac{5}{7}\right):\dfrac{4}{5}\)

\(=0:\dfrac{4}{5}\)

\(=0\)

6 tháng 12 2017

a,

\(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)

\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)

\(=1+1-0,5=1,5\)

b,

\(\dfrac{3}{7}\cdot19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)

\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{7}.\left(-14\right)=-6\)

c,

\(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)\)

\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)=-10:\left(-\dfrac{5}{7}\right)=14\)

d,

\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left(-\dfrac{2}{3}+\dfrac{3}{7}+\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left[\left(-\dfrac{2}{3}+\dfrac{-1}{3}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\right]:\dfrac{4}{5}\)

\(=\left(-1+1\right):\dfrac{4}{5}=0:\dfrac{4}{5}=0\)