\(\dfrac{7}{9}\)và\(\dfrac{...">
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13 tháng 6 2018

Dấu " / " là phân số nhé

a) 5/-4 . 16/25 + -5/4 . 9/25

= -5/4 . 16/25 + -5/4 . 9/25

= -5/4 . ( 16/25 + 9/25 )

= -5/4 . 1

= -5/4

b) 4 11/23 - 9/14 + 2 12/23 - 5/4

= 103/23 - 9/14 + 58/23 - 5/4

= 103/23 + 58/23 - 9/14 - 5/4

= 7 - 9/14 - 5/4

= 143/28

c) 2 13/27 - 7/15 + 3 14/27 - 8/15

= 67/27 - 7/15 + 95/27 - 8/15

= 67/27 + 95/27 - 7/15 - 8/15

= 6 - 7/15 - 8/15

= 5

a: \(=\dfrac{1}{4}\cdot\dfrac{12}{5}\cdot\dfrac{100}{7}\cdot\dfrac{49}{100}\)

\(=\dfrac{1}{4}\cdot\dfrac{12}{5}\cdot\dfrac{49}{7}=\dfrac{3}{5}\cdot7=\dfrac{21}{5}\)

b: \(=\dfrac{3}{8}+\dfrac{1}{8}\cdot\dfrac{3}{4}-\dfrac{5}{4}\)

\(=\dfrac{12}{32}+\dfrac{3}{32}-\dfrac{40}{32}=\dfrac{-25}{32}\)

c: \(=\dfrac{4}{9}\left(\dfrac{-13}{27}-\dfrac{14}{27}\right)-\dfrac{5}{9}=\dfrac{-4}{9}-\dfrac{5}{9}=-1\)

d: \(=\dfrac{2}{4}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{91\cdot95}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{91}-\dfrac{1}{95}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{92}{285}=\dfrac{46}{285}\)

a: \(=\dfrac{-28}{36}+\dfrac{15}{36}-\dfrac{26}{36}=\dfrac{-39}{36}=\dfrac{-13}{12}\)

b: \(=\dfrac{11}{9}\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)=\dfrac{11}{9}\cdot\dfrac{3}{4}=\dfrac{11}{12}\)

c: \(=15+\dfrac{9}{7}+6+\dfrac{2}{3}-5-\dfrac{5}{9}\)

\(=16+\dfrac{88}{63}=\dfrac{1096}{63}\)

d: \(=\dfrac{5}{6}-\dfrac{1}{3}+\dfrac{2}{18}\)

\(=\dfrac{15-6+2}{18}=\dfrac{11}{18}\)

25 tháng 1 2024

\(\dfrac{3}{2}\)\(\dfrac{-7}{5}\)\(\dfrac{4}{5}\)\(\dfrac{9}{11}\)\(\dfrac{0}{1}\)

2 = 2; 5 = 5; 11 = 11; 1 = 1

BCNN(2;5;11;1) = 2.5.11.1 = 110

Vậy mẫu chung nhỏ nhất của các phân số đã cho là: 110

 

Đề răng dài thế này thì tui giải từng câu hí

a) \(\dfrac{-7}{9}+\dfrac{5}{12}-\dfrac{13}{18}\left(MSC:36\right)\)

\(=\dfrac{-28}{36}+\dfrac{15}{36}-\dfrac{26}{36}\)

\(=\dfrac{-13}{36}-\dfrac{26}{36}\)

\(=\dfrac{-39}{36}\)

\(=\dfrac{13}{3}\)

b) \(\dfrac{11}{9}.\dfrac{15}{4}-\dfrac{11}{4}.\dfrac{7}{9}-\dfrac{11}{9}.\dfrac{5}{4}\)

\(=\left(\dfrac{15}{4}-\dfrac{11}{4}-\dfrac{5}{4}\right).\dfrac{11}{9}\)

\(=\left(1-\dfrac{5}{4}\right).\dfrac{11}{9}\)

\(=\left(\dfrac{4}{4}-\dfrac{5}{4}\right).\dfrac{11}{9}\)

\(=\dfrac{-1}{4}.\dfrac{11}{9}\)

\(=\dfrac{-11}{36}\)

17 tháng 4 2017

Giải bài 138 trang 58 SGK Toán 6 Tập 2 | Giải toán lớp 6

21 tháng 7 2018

\(a)\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\)

\(=\left(\dfrac{17}{6}+\dfrac{13}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{6}{12}\right)\)

\(=\left(\dfrac{153}{54}+\dfrac{78}{54}\right):\left(1\dfrac{-5}{12}\right)\)

\(=\dfrac{231}{54}:\dfrac{7}{12}\)

\(=\dfrac{198}{27}\)

21 tháng 7 2018

\(b)\dfrac{0,8\left(\dfrac{4}{5}:1,25\right)}{0,64-\dfrac{1}{25}}\)

\(=\dfrac{0,8\left(0,8:1,25\right)}{0,64-0,04}\)

\(=\dfrac{0,8.0,64}{0,6}\)

\(=\dfrac{0,512}{0,6}\)\(=\dfrac{64}{75}\)

16 tháng 5 2018

3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)

=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)

=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)

=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)

=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)

=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)

=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)

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

16 tháng 5 2018

4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)

=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)

=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)

=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)

=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)

=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)

=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)

=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)

=> \(3x=\dfrac{1}{9}\)

=> \(x=\dfrac{1}{9}:3\)

=> \(x=\dfrac{1}{27}\)

CÁCH 1 : A = \(\dfrac{235}{11}-\left(\dfrac{8}{5}+\dfrac{81}{11}\right)\)

A = \(\dfrac{235}{11}-\left(\dfrac{88}{55}+\dfrac{405}{55}\right)\)

A = \(\dfrac{235}{11}-\dfrac{493}{55}\)

A = \(\dfrac{1175}{55}+\dfrac{493}{55}\)

A = \(\dfrac{1668}{55}\)