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19 tháng 2 2019

a) Ta có:\(\dfrac{31}{67}>\dfrac{31}{73}\) (1)

\(\dfrac{31}{73}>\dfrac{29}{73}\) (2)

Từ (1) và (2) \(\Rightarrow\) \(\dfrac{31}{67}>\dfrac{31}{73}>\dfrac{29}{73}\)

\(\Rightarrow\dfrac{31}{67}>\dfrac{29}{73}\)

Vậy:...............

3 tháng 7 2017

a)\(12< 13;49>47\)

\(\Rightarrow\dfrac{12}{49}< \dfrac{13}{47}\)

b)\(\dfrac{64}{85}>\dfrac{43}{85}\Rightarrow\dfrac{64}{85}>\dfrac{1}{2}\)

\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)

\(\Rightarrow\dfrac{17}{35}< \dfrac{64}{85}\)

c) \(\dfrac{19}{31}>\dfrac{16}{31}\Rightarrow\dfrac{19}{31}>\dfrac{1}{2}\)

\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)

\(\Rightarrow\dfrac{17}{35}< \dfrac{19}{31}\)

d)

\(1-\dfrac{67}{77}=\dfrac{10}{77}\)

\(1-\dfrac{73}{83}=\dfrac{10}{83}\)

\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)

e)\(1-\dfrac{456}{461}=\dfrac{5}{461}\)

\(1-\dfrac{123}{128}=\dfrac{5}{128}\)

\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)

3 tháng 7 2017

\(a,\dfrac{12}{49}< \dfrac{12}{47}< \dfrac{13}{47}\Rightarrow\dfrac{12}{49}< \dfrac{12}{47}\)

b, Ta có: \(\dfrac{17}{35}=\dfrac{51}{105}\)

\(\dfrac{64}{85}>\dfrac{64}{105}>\dfrac{51}{105}\Rightarrow\dfrac{64}{85}>\dfrac{51}{105}\) hay \(\dfrac{64}{85}>\dfrac{17}{85}\)

c,\(\dfrac{19}{31}>\dfrac{17}{31}>\dfrac{17}{35}\Rightarrow\dfrac{19}{31}>\dfrac{17}{35}\)

d, \(\dfrac{67}{77}+\dfrac{10}{77}=1\)

\(\dfrac{73}{83}+\dfrac{10}{83}=1\)

\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)

e, \(\dfrac{456}{461}+\dfrac{5}{461}=1\)

\(\dfrac{123}{128}+\dfrac{5}{128}=1\)

\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)

26 tháng 3 2017

Ta có: \(\dfrac{20}{39}>\dfrac{20}{41}>\dfrac{18}{41}\left(1\right)\)

\(\dfrac{22}{27}>\dfrac{22}{29}\left(2\right)\)

\(\dfrac{18}{43}=1-\dfrac{25}{43};\dfrac{14}{39}=1-\dfrac{25}{39}\)

\(\dfrac{25}{43}< \dfrac{25}{39}\Rightarrow1-\dfrac{25}{43}>1-\dfrac{25}{39}\Rightarrow\dfrac{18}{43}>\dfrac{14}{39}\left(3\right)\)

Từ \(\left(1\right);\left(2\right);\left(3\right)\) ta suy ra : A>B

6 tháng 4 2017

bn là nam hay nữ zợ

11 tháng 7 2017

các bạn giúp mk vs

mk đg cần gấp khocroi

16 tháng 8 2018

b, Ta có: \(\dfrac{58}{53}>1>\dfrac{36}{55}\)

hay \(\dfrac{58}{53}>\dfrac{36}{55}\)

\(\Rightarrow0-\dfrac{58}{53}< 0-\dfrac{36}{55}\)

\(\Rightarrow\dfrac{-58}{53}< \dfrac{-36}{55}\)

1 tháng 7 2017

a, \(\dfrac{1998}{1999}\) < \(\dfrac{1999}{2000}\)

b, \(\dfrac{47}{15}>\dfrac{29}{35}\)

c, \(\dfrac{12}{25}< \) \(\dfrac{25}{49}\)

27 tháng 2 2019

1 )Ta có

\(M=\left(\dfrac{1}{2^2}-1\right)\cdot\left(\dfrac{1}{3^2}-1\right)\cdot\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right).....\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{4}{3}\cdot\dfrac{-3}{4}\cdot\dfrac{5}{4}\cdot\cdot\cdot\cdot\dfrac{-99}{100}\cdot\dfrac{101}{100}\)

\(=\dfrac{-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot3\cdot\left(-4\right)\cdot4\cdot\left(-5\right)\cdot5....\cdot\left(-100\right)\cdot100\cdot101}{2^2\cdot3^2\cdot4^2....\cdot100^2}\)

\(=-\dfrac{101}{200}< \dfrac{1}{2}\)

2 ) Số phân số của biểu thức B là 180 phân số

Ta có

\(\dfrac{1}{20}>\dfrac{1}{200};\dfrac{1}{21}>\dfrac{1}{200};\dfrac{1}{22}>\dfrac{1}{200};....;\dfrac{1}{199}>\dfrac{1}{200}\)

\(\Rightarrow B=\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{200}>\dfrac{1}{200}\cdot180=\dfrac{9}{10}\)

25 tháng 7 2017

\(\dfrac{x-13}{87}+\dfrac{x-27}{73}+\dfrac{x-67}{83}+\dfrac{x-73}{27}=4\)

<=>\(\dfrac{x-13}{87}-1+\dfrac{x-27}{73}-1+\dfrac{x-67}{83}-1+\dfrac{x-73}{27}-1=0\)

<=>\(\dfrac{x-100}{87}+\dfrac{x-100}{73}+\dfrac{x-100}{83}+\dfrac{x-100}{27}=0\)

<=>\(\left(x-100\right)\left(\dfrac{1}{87}+\dfrac{1}{73}+\dfrac{1}{83}+\dfrac{1}{27}\right)=0\)

Do \(\dfrac{1}{87}+\dfrac{1}{73}+\dfrac{1}{83}+\dfrac{1}{27}>0\)

=>x-100=0

<=>x=100

25 tháng 7 2017

sửa chỗ :\(\dfrac{x-67}{33}\)