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\(5\frac{1}{2}>5\frac{1}{4}\)
\(12\frac{7}{12}< 12\frac{5}{8}\)
\(6\frac{5}{7}< 12\frac{5}{8}\)
\(9\frac{6}{7}>8\frac{7}{8}\)
#Yuki
`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) Vì 4 < 5 nên \(\frac{3}{4}>\frac{3}{5}\)
b) Vì 4 < 5 nên \(\frac{4}{7}< \frac{5}{7}\)
c) Ta có :
\(\frac{2}{7}=\frac{2\times5}{7\times5}=\frac{10}{35}\)
\(\frac{4}{5}=\frac{4\times7}{5\times7}=\frac{28}{35}\)
Vì 10 < 28 nên \(\frac{10}{35}< \frac{28}{35}\)hay \(\frac{2}{7}< \frac{4}{5}\)
d) Ta có :
\(\frac{8}{9}=\frac{8\times10}{9\times10}=\frac{80}{90}\)
\(\frac{9}{10}=\frac{9\times9}{10\times9}=\frac{81}{90}\)
Vì 80 < 81 nên \(\frac{80}{90}< \frac{81}{90}\)hay \(\frac{8}{9}< \frac{9}{10}\)
e) Ta có :
\(\frac{10}{9}=\frac{10\times2}{9\times2}=\frac{20}{18}\)
Vì 19 < 20 nên \(\frac{19}{18}< \frac{20}{18}\)hay \(\frac{19}{18}< \frac{10}{9}\)
f) Ta có :
\(\frac{8}{3}=\frac{8\times4}{3\times4}=\frac{32}{12}\)
\(\frac{5}{6}=\frac{5\times2}{6\times2}=\frac{10}{12}\)
Vì 32 > 10 nên \(\frac{32}{12}>\frac{10}{12}\)hay \(\frac{8}{3}>\frac{5}{6}\)
a) Ta có :
1912 / 1917 =0,997..
3035/3040 = 0,998...
Vì 0,997...<0,998... nên 1912/1917 < 3035/3040
b)Ta có: 20/18:40/38 = 760/720
Vì 760/720 > 1 nên 20/18 > 40/38
c) Ta có :
Vì 3/7 < 1 ; 7/4 > 1
nên 3/7 < 7/4
d) Ta có : 1/2 = 20/40 ; 2/5 = 16/40 ; 4/8 = 20/40
Vì 10/40 < 16/40 < 20/40 = 20/40
Nên 10/40 < 2/5 < 1/2 = 4/8
e) 8/2 = 4 ; 5/3 = 1,66... ; 6/5 = 1,2 ; 1/6 = 0,16...
Vì 0,16...< 1, 2 < 1,66...< 4
Nên 1/6 < 6/5 < 5/3 / 8/2
`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`
Bài 4
35/85 = 7/17
36/108 = 1/3
25/100 = 1/4
39/52 = 3/4
Bài 8
a) 9/8 và 7/12
= 8×3=24 ; 12×2=24
=>9/8 =27/24
=> 7/12 ; 14/24
b) 3/20 và 4/15
=20×3=60 ; 15×4=60
=> 9/60 ; 16/60
Bài 9
a) \(\frac{3}{8},\frac{15}{8},\frac{9}{8},\frac{7}{8}\)
Từ lớn -> bé:
=>\(\frac{15}{8},\frac{9}{8},\frac{7}{8},\frac{3}{8}\)
b) \(\frac{4}{15},\frac{3}{5},\frac{8}{45},\frac{7}{15}=\frac{12}{45},\frac{27}{45},\frac{8}{45},\frac{21}{45}\)
Từ lớn -> bé:
=> \(\frac{3}{5},\frac{7}{15},\frac{4}{15},\frac{8}{45}\)
c) \(\frac{3}{8},\frac{4}{5},\frac{47}{40},\frac{9}{4}=\frac{15}{40},\frac{32}{40},\frac{47}{40},\frac{90}{40}\)
Từ lớn -> bé:
=>\(\frac{9}{4},\frac{47}{40},\frac{4}{5},\frac{3}{8}\)
Bài 10
a, Ta có
`x/15 < 4/15`
` <=> x < 4`
` <=> x ∈ {1 ; 2 ; 3}`
b, Ta có
`5/9 > x/9`
` <=> 5 > x`
` <=> x ∈ {1 ; 2 ; 3 ; 4}`
c, Ta có
`1 <x/8 < 11/8`
` <=> 8/8 < x/8 < 11/8`
` <=> 8 < x <11`
` <=> x ∈ {9 ; 10}`
a; Cách một:
\(\dfrac{2}{9}\) = \(\dfrac{2\times2}{9\times2}\) = \(\dfrac{4}{18}\) < \(\dfrac{4}{10}\) Vậy \(\dfrac{2}{9}\) < \(\dfrac{4}{10}\)
\(\dfrac{4}{9}\) = \(\dfrac{4\times3}{9\times3}\) = \(\dfrac{12}{27}\); \(\dfrac{6}{10}\) = \(\dfrac{6\times2}{10\times2}\) = \(\dfrac{12}{20}\)
Vì \(\dfrac{12}{27}\) < \(\dfrac{12}{20}\) vậy \(\dfrac{4}{9}\) < \(\dfrac{12}{20}\)
\(\dfrac{3}{8}\) = \(\dfrac{3\times4}{8\times4}\) = \(\dfrac{12}{24}\); \(\dfrac{4}{7}\) = \(\dfrac{4\times3}{7\times3}\) = \(\dfrac{12}{21}\)
Vậy \(\dfrac{3}{8}\) < \(\dfrac{4}{7}\)
\(\dfrac{5}{9}\) = \(\dfrac{5\times7}{9\times7}\) = \(\dfrac{35}{63}\); \(\dfrac{7}{10}\) = \(\dfrac{7\times5}{10\times5}\) = \(\dfrac{35}{50}\)
Vì \(\dfrac{35}{63}\) < \(\dfrac{35}{50}\) vậy \(\dfrac{5}{9}\) < \(\dfrac{7}{10}\)
Cách hai:
a; \(\dfrac{2}{9}\) = \(\dfrac{2\times10}{9\times10}\) = \(\dfrac{20}{90}\); \(\dfrac{4}{10}\) = \(\dfrac{4\times9}{10\times9}\) = \(\dfrac{36}{90}\)
Vì \(\dfrac{20}{90}\) < \(\dfrac{36}{90}\) vậy \(\dfrac{2}{9}\) < \(\dfrac{4}{10}\)
b; \(\dfrac{4}{9}\) = \(\dfrac{4\times10}{9\times10}\) = \(\dfrac{40}{90}\); \(\dfrac{6}{10}\) = \(\dfrac{6\times9}{10\times9}\) = \(\dfrac{54}{90}\)
Vì \(\dfrac{40}{90}\) < \(\dfrac{54}{90}\) vậy \(\dfrac{4}{9}\) < \(\dfrac{6}{10}\)
c; \(\dfrac{3}{8}\) = \(\dfrac{3\times7}{8\times7}\) = \(\dfrac{21}{56}\); \(\dfrac{4}{7}\) = \(\dfrac{4\times8}{7\times8}\) = \(\dfrac{32}{56}\)
Vì \(\dfrac{21}{56}\) < \(\dfrac{32}{56}\) vậy \(\dfrac{3}{8}\) < \(\dfrac{4}{7}\)
d; \(\dfrac{5}{9}\) = \(\dfrac{5\times10}{9\times10}\) = \(\dfrac{50}{90}\); \(\dfrac{7}{10}\) = \(\dfrac{7\times9}{10\times9}\) = \(\dfrac{63}{90}\)
Vì \(\dfrac{50}{90}\) < \(\dfrac{63}{90}\) vậy \(\dfrac{5}{9}\) < \(\dfrac{7}{10}\)
a, >
b, <
c, >
A 3/7 > 2/5
B 5/9 < 5/8
C 8/7 > 7/8
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