\(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\)

Ta có :

+) \(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}\)

+) \(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}\)

\(\Leftrightarrow S< \dfrac{1}{5}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{40}+\dfrac{1}{40}\)

\(\Leftrightarrow S< \dfrac{1}{2}\)

Vậy,,,

Ta có: \(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)

\(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{2}{40}=\dfrac{1}{20}\)

Do đó: \(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{4}+\dfrac{1}{20}=\dfrac{6}{20}=\dfrac{3}{10}\)

\(\Leftrightarrow\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{3}{10}+\dfrac{1}{5}=\dfrac{3}{10}+\dfrac{2}{10}=\dfrac{1}{2}\)

hay \(S< \dfrac{1}{2}\)(đpcm)

a: 8/9<1<9/3

b: 8/4=2=6/3

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=

a) 8/9 < 1 < 9/3

b) 8/4 = 2 = 6/3

a) \(\dfrac{5}{9}< \dfrac{7}{9}\)

b) \(\dfrac{7}{6}>\dfrac{6}{6}\)

c) \(\dfrac{3}{14}< \dfrac{5}{14}\)

d) \(\dfrac{5}{8}< \dfrac{9}{8}\)

7/9<8/9

Có : 7<8

\(\Rightarrow\dfrac{7}{9}< \dfrac{8}{9}\)

\(\dfrac{7}{9}< \dfrac{8}{9}\)

`3/7-2/5`

`=1/35>0`

`=>3/7>2/5`

`b,9>8`

`=>1/9<1/8`

`=>5/9<5/8`

`d,8/7>1`

`7/8<1`

`=>8/7>7/8`

a) Ta có: \(\dfrac{3}{7}=\dfrac{15}{35}\)

\(\dfrac{2}{5}=\dfrac{14}{35}\)

Do đó: \(\dfrac{3}{7}>\dfrac{2}{5}\)

b) Ta có: 9>8

nên \(\dfrac{5}{9}< \dfrac{5}{8}\)

c) Ta có: \(\dfrac{8}{7}>1\)

\(1>\dfrac{7}{8}\)

Do đó: \(\dfrac{8}{7}>\dfrac{7}{8}\)

a)\(\dfrac{-8}{9}< \dfrac{-7}{9}\\ \dfrac{6}{7}< \dfrac{11}{10}\)

a) Vì -8<-7 nên \(\dfrac{-8}{9}< \dfrac{-7}{9}\)

b) Ta có: \(\dfrac{6}{7}< 1;\dfrac{11}{10}>1\)

nên \(\dfrac{6}{7}< \dfrac{11}{10}\)

Giải:

a) -8/9 < -7/9

b) Ta có: 6/7 = 60/70

                 11/10 = 77/70

Vì 60/70 < 77/70 nên :

    6/7 < 11/10

8/9<9/9

8/3>8/9

< ; >

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> =)

\(7\dfrac{4}{5}và9\dfrac{1}{2}\\ Tacó:7< 9\\ \Rightarrow7\dfrac{4}{5}< 9\dfrac{1}{2}\\ 7\dfrac{1}{6}và3\dfrac{4}{5}\\ Tacó:7>3\\ \Rightarrow7\dfrac{1}{6}>3\dfrac{4}{5}\)

Câu cuối không phải hỗn số

`a)1<3`

`=>1/5<3/5`

`b)21>9`

`=>8/21<8/9`

`c)3/5<5/5=1`

`d)7/5>5/5=1`

a)3/5>1/5 b)8/21<8/9 c)3/5<1 d)7/5>1

a 1/5<3/5

b 8/21<8/9

c 3/5<1

d 7/5>1