\(\dfrac{11}{120}\) và \(\dfrac{7}{40}\)

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10 tháng 1 2024

\(\dfrac{11}{120}\) và \(\dfrac{7}{40}\)

\(\Rightarrow BCNN\left(120,40\right)=MC=120.\)

\(\Rightarrow\) Ta có:

\(\dfrac{7}{40}=\dfrac{7.3}{40.3}=\dfrac{21}{120};\dfrac{11}{120}\)

Mà \(\dfrac{11}{120}< \dfrac{21}{120}\Rightarrow\dfrac{11}{120}< \dfrac{7}{40}\) 

10 tháng 1 2024

Đề yêu cầu so sánh hay quy đồng hay làm gì với 2 phân số này em?

16 tháng 4 2017

a)\(\dfrac{11}{120};\dfrac{21}{120}\)

b)\(\dfrac{312}{1898};\dfrac{876}{1898}\)

c)\(\dfrac{28}{120};\dfrac{26}{120};\dfrac{-27}{120}\)

d)\(\dfrac{51}{180};\dfrac{-50}{180};\dfrac{-128}{180}\)

8 tháng 3 2017

\(A=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{-5}{7}\)

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

\(B=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

\(=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

\(=\dfrac{7}{10}.\dfrac{5}{28}.20=\dfrac{5}{2}.\)

8 tháng 3 2017

Ta có :

A = \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)

= \(\dfrac{5}{7}.\dfrac{-2}{11}-\dfrac{5}{7}.\dfrac{9}{11}+\dfrac{5}{7}+1\)

= \(\left(\dfrac{5}{7}.\dfrac{-2}{11}-\dfrac{5}{7}.\dfrac{9}{11}+\dfrac{5}{7}\right)+1\)

= \(\dfrac{5}{7}.\left(\dfrac{-2}{11}-\dfrac{9}{11}+1\right)+1\)

= \(\dfrac{5}{7}.0+1\)

= 1

B = \(0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

= \(\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

= \(\left(\dfrac{7}{10}.20\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).\dfrac{5}{28}\)

= 14.1.\(\dfrac{5}{28}\)

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

Vậy A = 1

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

9 tháng 4 2017

a) \(\dfrac{37}{40}-0,64\\ =\dfrac{37}{40}-\dfrac{16}{25}\\ =\dfrac{185}{200}-\dfrac{128}{200}\\ =\dfrac{57}{200}\)

9 tháng 4 2017

b) \(130\dfrac{25}{28}-120\dfrac{12}{35}\\ =\dfrac{3665}{28}-\dfrac{4212}{35}\\ =\dfrac{18325}{140}-\dfrac{16848}{140}\\ =\dfrac{211}{20}\)

17 tháng 5 2018

Giải:

a) \(A=\dfrac{5}{13}.\dfrac{5}{7}+\dfrac{-20}{41}+\dfrac{5}{13}+\dfrac{-21}{41}\)

\(\Leftrightarrow A=\dfrac{5}{13}.\dfrac{5}{7}+\dfrac{5}{13}+\dfrac{-21}{41}+\dfrac{-20}{41}\)

\(\Leftrightarrow A=\dfrac{5}{13}\left(\dfrac{5}{7}+1\right)+\dfrac{-41}{41}\)

\(\Leftrightarrow A=\dfrac{5}{13}.\dfrac{12}{7}+\left(-1\right)\)

\(\Leftrightarrow A=\dfrac{60}{91}+\left(-1\right)=-\dfrac{31}{91}\)

Vậy ...

b) \(B=\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{12}{11}-\dfrac{5}{7}.\dfrac{7}{11}\)

\(\Leftrightarrow B=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{12}{11}-\dfrac{7}{11}\right)\)

\(\Leftrightarrow B=\dfrac{5}{7}.\dfrac{7}{11}\)

\(\Leftrightarrow B=\dfrac{5}{11}\)

Vậy ...

c) \(C=\dfrac{-2}{3}+\dfrac{-5}{7}+\dfrac{2}{3}+\dfrac{-2}{7}\)

\(\Leftrightarrow C=\left(\dfrac{-2}{3}+\dfrac{2}{3}\right)+\left(\dfrac{-2}{7}+\dfrac{-5}{7}\right)\)

\(\Leftrightarrow C=0+\left(-1\right)=-1\)

Vậy ...

a)\(\dfrac{-5}{9}+\dfrac{5}{9}:\left(1\dfrac{2}{3}-2\dfrac{1}{12}\right)\)

\(=\dfrac{-5}{9}+\dfrac{5}{9}:\left(\dfrac{20}{12}-\dfrac{25}{12}\right)\)

\(=\dfrac{-5}{9}+\dfrac{5}{9}\cdot\dfrac{-12}{5}\)

\(=\dfrac{-5}{9}+\dfrac{-12}{9}\)

\(=\dfrac{-17}{9}\)

b) \(\dfrac{-7}{25}\cdot\dfrac{11}{13}+\dfrac{-7}{25}\cdot\dfrac{2}{13}-\dfrac{18}{25}\)

\(=\dfrac{-7}{25}\cdot\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}\)

\(=\dfrac{-7}{25}\cdot1-\dfrac{18}{25}\)

\(=\dfrac{-25}{25}\)

\(=-1\)

c) \(\left(\dfrac{5}{7}\cdot0,6-5:3\dfrac{1}{2}\right)\cdot\left(40\%-1,4\right)\cdot\left(-2\right)^3\)

\(=\left(\dfrac{3}{7}-\dfrac{10}{7}\right)\cdot\left(0,4-1,4\right)\cdot\left(-8\right)\)

\(=\left(-1\right)\cdot\left(-1\right)\cdot\left(-8\right)\)

\(=-8\)

10 tháng 6 2018

Giải:

a)

\(\dfrac{7}{48}=\dfrac{105}{720};\)

\(\dfrac{11}{72}=\dfrac{110}{720};\)

\(\dfrac{17}{120}=\dfrac{102}{720}\)

\(102< 105< 110\)

\(\Leftrightarrow\dfrac{102}{720}< \dfrac{105}{720}< \dfrac{110}{720}\)

\(\Leftrightarrow\dfrac{17}{120}< \dfrac{7}{48}< \dfrac{11}{72}\)

Vậy ...

b) \(\dfrac{31}{49}=\dfrac{60140}{95060};\)

\(\dfrac{62}{97}=\dfrac{60760}{95060};\)

\(\dfrac{93}{140}=\dfrac{63147}{95060}\)

\(60140< 60760< 63147\)

\(\Leftrightarrow\dfrac{60140}{95060}< \dfrac{60760}{95060}< \dfrac{63147}{95060}\)

\(\Leftrightarrow\dfrac{31}{49}< \dfrac{62}{97}< \dfrac{93}{140}\)

Vậy ...

11 tháng 6 2018

a ) \(\dfrac{7}{48}\) = \(\dfrac{105}{720}\)

\(\dfrac{11}{72}\) = \(\dfrac{110}{720}\)

\(\dfrac{17}{120}\) = \(\dfrac{102}{720}\)

Vì 102 < 105 < 110

\(\Leftrightarrow\) \(\dfrac{102}{720}\) < \(\dfrac{105}{720}\) < \(\dfrac{110}{720}\)

\(\Leftrightarrow\) \(\dfrac{17}{120}\) < \(\dfrac{7}{48}\) < \(\dfrac{11}{72}\)

Vậy .....................

( k cho tớ nha . Tớ chỉ bt lm phần a )

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

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

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

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

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

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

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

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

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