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10 tháng 4 2017

\(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\\ =\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{-3\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\\ =\dfrac{2}{-3}\\ =-\dfrac{2}{3}\)

20 tháng 8 2017

a, \(4\dfrac{5}{37}\)-\(3\dfrac{4}{5}\)+ \(8\dfrac{15}{29}\)- \(3\dfrac{5}{37}\)+ \(6\dfrac{14}{29}\)

=(\(4\dfrac{5}{37}\)-\(3\dfrac{5}{37}\))+(\(8\dfrac{15}{29}\)+\(6\dfrac{14}{29}\))-\(3\dfrac{4}{5}\)

=(4-3)+(\(\dfrac{5}{37}\)-\(\dfrac{5}{37}\))+(8+6)+(\(\dfrac{15}{29}\)+\(\dfrac{14}{29}\))-3\(\dfrac{4}{5}\)

=1+ 15-\(3\dfrac{4}{5}\)=13-\(\dfrac{4}{5}\)=\(\dfrac{61}{5}\)

b, 60\(\dfrac{7}{13}\)+ 50\(\dfrac{8}{13}\)-11\(\dfrac{2}{13}\)

=(60+50-11)+(\(\dfrac{7}{13}\)+ \(\dfrac{8}{13}\)-\(\dfrac{2}{13}\))

=99+1=100

c, đáp án bằng \(\dfrac{-2}{3}\). bạn tự tính nha

a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)

a: \(=\left(\dfrac{28}{42}+\dfrac{12}{42}-\dfrac{3}{42}\right):\left(\dfrac{-28}{28}-\dfrac{12}{28}+\dfrac{3}{28}\right)\)

\(=\dfrac{37}{42}:\dfrac{-37}{28}=\dfrac{-28}{42}=-\dfrac{2}{3}\)

b: \(=\dfrac{2+8+18+32+50}{12+48+108+192+300}=\dfrac{110}{660}=\dfrac{1}{6}\)

11 tháng 4 2017

a) \(\dfrac{1}{2}.\left(\dfrac{2}{9}+\dfrac{3}{7}-\dfrac{5}{27}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{41}{63}-\dfrac{5}{27}\right)\)

\(=\dfrac{1}{2}.\dfrac{88}{189}\)

\(=\dfrac{44}{189}\)

b) \(\left(\dfrac{-5}{28}+1,75+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)

\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)

\(=\dfrac{9}{5}:\left(-3\dfrac{9}{20}\right)\)

\(=\dfrac{9}{5}:\dfrac{-69}{20}\)

\(=\dfrac{-12}{23}\)

c) \(\dfrac{1}{3}.\dfrac{5}{7}-\dfrac{7}{27}.\dfrac{36}{14}\)

\(=\dfrac{5}{21}-\dfrac{7}{27}.\dfrac{36}{14}\)

\(=\dfrac{5}{21}-\dfrac{2}{3}\)

\(=\dfrac{-3}{7}\)

d) \(70,5-528:\dfrac{15}{2}\)

\(=70,5-\dfrac{352}{5}\)

\(=\dfrac{1}{10}\)

11 tháng 4 2017

em không trả lời được câu hỏi của chị nhưng chị có thể giúp em đăng bài toán lên bằng cách nào không

18 tháng 7 2017

a) \(6\dfrac{5}{7}-\left(1\dfrac{3}{4}+2\dfrac{5}{7}\right)\)

\(=6\dfrac{5}{7}-1\dfrac{3}{4}-2\dfrac{5}{7}\)

\(=\left(6\dfrac{5}{7}-2\dfrac{5}{7}\right)-1\dfrac{3}{4}\)

\(=4-1\dfrac{3}{4}\)

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

18 tháng 7 2017

b) \(7\dfrac{5}{11}-\left(2\dfrac{3}{7}+3\dfrac{5}{11}\right)\)

\(=7\dfrac{5}{11}-2\dfrac{3}{7}-3\dfrac{5}{11}\)

\(=\left(7\dfrac{5}{11}-3\dfrac{5}{11}\right)-2\dfrac{3}{7}\)

\(=4-2\dfrac{3}{7}\)

\(=2\dfrac{3}{7}\)

4 tháng 6 2017

a) \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

\(=\dfrac{1135}{23}-\left(\left(5+14\right)+\left(\dfrac{7}{32}+\dfrac{8}{23}\right)\right)\)

\(=\dfrac{1135}{23}-\left(19+\dfrac{417}{736}\right)\)

\(=\dfrac{1135}{23}-19\dfrac{417}{736}\)

\(=\dfrac{1135}{23}-\dfrac{14401}{736}\)

\(=\dfrac{953}{32}\)

b) \(-\dfrac{3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)

\(=-\dfrac{1}{7}\cdot\dfrac{5}{3}-\dfrac{4}{3}\cdot\dfrac{1}{7}+\dfrac{17}{7}\)

\(=-\dfrac{5}{21}-\dfrac{4}{21}+\dfrac{17}{7}\)

\(=2\)

c) \(0,7\cdot2\dfrac{2}{3}\cdot20\cdot0,375\cdot\dfrac{5}{28}\)

\(=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)

\(=2\cdot\dfrac{5}{4}\)

\(=\dfrac{5}{2}\)

d) \(\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)

\(=\left(9\cdot\dfrac{3}{8}+\dfrac{303030}{69264}\right)+\dfrac{403}{100}\)

\(=\left(\dfrac{27}{8}+\dfrac{35}{8}\right)+\dfrac{403}{100}\)

\(=\dfrac{31}{4}+\dfrac{403}{100}\)

\(=\dfrac{589}{50}\)

P/s: Đánh dấu phẩy, dấu chấm (dấu nhân) cần rõ ràng (vì dấu chấm người ta sẽ hiểu là dấu nhân thay vì hiểu là dấu phẩy)

5 tháng 6 2017

a) \(49\dfrac{8}{23}\)- \(\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

= \(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)+5\dfrac{7}{32}\)

=35+\(5\dfrac{7}{32}\)

=\(\dfrac{1287}{32}\)

b)\(-\dfrac{3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)

=\(\left[\left(\dfrac{-3}{7}\right).\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\right]+2\dfrac{3}{7}\)

=\(\left[\left(\dfrac{-3}{7}\right).\dfrac{9}{9}\right]+2\dfrac{3}{7}\)

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

=\(\left(\dfrac{-3}{7}\right)+2\dfrac{3}{7}\)

=2

c) 0,7.\(2\dfrac{2}{3}\).20.0.375.\(\dfrac{5}{28}\)

=0 (Vì có một thừ số là 0 nên nguyên cả tích là 0)

d)\(\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)

=17+4,03

=21,03

1:-5/14=-50/140

3/20=21/140

9/70=18/140

2: -55/132=-5/12=-35/84

10/42=20/84

-3/28=-9/84

3: 7/10=231/330

1/33=10/330

17 tháng 4 2017

Thay dấu ba chấm bởi x rồi tìm x.

Chẳng hạn:

\(a) \) \(\dfrac{7}{9}-\dfrac{x}{3}=\dfrac{1}{9}\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{7}{9}-\dfrac{1}{9}\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{6}{9}=\dfrac{2}{3}\)

Vậy x = 2

Đáp số:

a) x = 2

b) x = 3

c) x = 7

d) x =19.

3 tháng 5 2018

Giải bà i 64 trang 34 SGK Toán 6 Tập 2 | Giải toán lá»p 6

18 tháng 6 2018

\(M=\dfrac{8}{3}\cdot\dfrac{2}{5}\cdot\dfrac{3}{8}\cdot10\cdot\dfrac{19}{92}\\ =\dfrac{8\cdot2\cdot3\cdot10\cdot19}{3\cdot5\cdot8\cdot92}\\ =\dfrac{8\cdot2\cdot3\cdot2\cdot5\cdot19}{3\cdot5\cdot8\cdot2\cdot2\cdot23}\\ =\dfrac{19}{23}\)

\(N=\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\\ =\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot0\\ =0\)