\(\dfrac{47}{95}\) và \(\dfrac{35}{69}\)

\(\dfrac{47}{95}< \dfrac{1}{2}\) và \(\dfrac{35}{69}>\dfrac{1}{2}\)

Vậy \(\dfrac{47}{95}< \dfrac{35}{69}\)

\(\dfrac{53}{103}\) và \(\dfrac{71}{145}\)

\(\dfrac{53}{103}>\dfrac{1}{2}\) và \(\dfrac{71}{145}< \dfrac{1}{2}\)

Vậy \(\dfrac{53}{103}>\dfrac{71}{145}\)

\(\dfrac{2009}{2010}\) và \(\dfrac{2005}{2006}\)

\(1-\dfrac{2009}{2010}=\dfrac{1}{2010}\) và \(1-\dfrac{2005}{2006}=\dfrac{1}{2006}\)

Vậy \(\dfrac{2009}{2010}>\dfrac{2005}{2006}\)

\(\dfrac{783}{901}\) và \(\dfrac{738}{915}\)

\(\dfrac{738}{915}< \dfrac{783}{915}< \dfrac{783}{901}\)

Vậy \(\dfrac{783}{901}>\dfrac{738}{915}\)

nhanh giúp mình với ạ

cảm ơn bạn nhiều

`a)75<77`

`=>1/75>1/77`

`=>37/75>37/77>35/77`

`b)7779>7569`

`=>1/7779<1/7569`

`=>3734/7569>3734/7779>2099/7779`

\(\sqrt{Ai}giúpmkvớiak\)

`M=(10^25+1)/(10^26+1)`

`=>10M=(10^26+10)/(10^26+1)=1+9/(10^26+1)``

`CMTT:10N=1+9/(10^27+1)`

Vì `1/(10^26+1)>1/(10^27+1)`

`=>9/(10^26+1)>9/(10^27+1)`

`=>1+9/(10^26+1)>1+9/(10^27+1)`

`=>10M>10N=>M>N`

\(\dfrac{14}{21};\dfrac{60}{72}\)

\(\dfrac{60}{72}=\dfrac{5}{6}=\dfrac{5\cdot21}{6\cdot21}=\dfrac{105}{126}\)

\(\dfrac{14}{21}=\dfrac{14\cdot6}{21\cdot6}=\dfrac{84}{126}\)

\(\dfrac{84}{123}< \dfrac{105}{126}\Rightarrow\dfrac{14}{21}< \dfrac{60}{72}\)

\(\dfrac{38}{133};\dfrac{129}{344}\)

\(\dfrac{38}{133}=\dfrac{19}{7}=\dfrac{19\cdot8}{7\cdot8}=\dfrac{152}{56}\)

\(\dfrac{129}{344}=\dfrac{3}{8}=\dfrac{3\cdot7}{8\cdot7}=\dfrac{21}{56}\)

\(\dfrac{152}{56}>\dfrac{21}{56}\Rightarrow\dfrac{38}{133}>\dfrac{129}{344}\)

1,

Ta có:
\(\dfrac{73}{75}=1-\dfrac{2}{75}\)

\(\dfrac{77}{79}=1-\dfrac{2}{79}\)
So sánh phân số \(\dfrac{2}{75}\) và \(\dfrac{2}{79}\)
Vì \(75< 79\) nên \(\dfrac{1}{75}>\dfrac{1}{79}\)
Vậy \(1-\dfrac{2}{75}< 1-\dfrac{2}{79}\)
Hay \(\dfrac{73}{75}< \dfrac{77}{79}\)

2,

Vì \(\dfrac{53}{100}>\dfrac{47}{100}>\dfrac{47}{106}\) nên \(\dfrac{53}{100}>\dfrac{47}{106}\)

3,

Ta có:
\(\dfrac{81}{79}=1+\dfrac{2}{79}\)

\(\dfrac{65}{63}=1+\dfrac{2}{63}\)
So sánh phân số \(\dfrac{2}{79}\) và \(\dfrac{2}{63}\)
Vì \(79>63\) nên \(\dfrac{81}{79}< \dfrac{65}{63}\)
Hay \(\Rightarrow1+\dfrac{2}{79}< 1+\dfrac{2}{63}\)
Vậy \(\dfrac{81}{79}< \dfrac{65}{63}\)

4,

\(\dfrac{48}{47}>1>\dfrac{84}{85}\)

Vậy \(\dfrac{48}{47}>\dfrac{84}{85}\)

giúp mình với ạ

cảm ơn ạ

`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

\(a,\dfrac{-15}{17}=-1+\dfrac{2}{17}\\ -\dfrac{19}{21}=-1+\dfrac{2}{21}\\ Vì:\dfrac{2}{17}>\dfrac{2}{21}\Rightarrow-1+\dfrac{2}{17}>-1+\dfrac{2}{21}\Rightarrow-\dfrac{15}{17}>-\dfrac{19}{21}\\ b,-\dfrac{24}{35}=-1+\dfrac{11}{35};-\dfrac{19}{30}=-1+\dfrac{11}{30}\\ Vì:\dfrac{11}{35}< \dfrac{11}{30}\Rightarrow-1+\dfrac{11}{35}< -1+\dfrac{11}{30}\\ \Rightarrow-\dfrac{24}{35}< -\dfrac{19}{30}\)

c) \(\dfrac{27}{26}\)và\(\dfrac{38}{37}\)

Ta có: \(\dfrac{27}{26}=1+\dfrac{1}{26}\); \(\dfrac{38}{37}=1+\dfrac{1}{37}\)

Vì \(\dfrac{1}{26}>\dfrac{1}{37}\) nên \(\dfrac{27}{26}>\dfrac{38}{37}\)

c: \(\dfrac{27}{26}-1=\dfrac{1}{26}\)

\(\dfrac{38}{37}-1=\dfrac{1}{37}\)

mà \(\dfrac{1}{26}>\dfrac{1}{37}\)

nên \(\dfrac{27}{26}>\dfrac{38}{37}\)

a) \(\dfrac{23}{24}< 1\)

\(\dfrac{24}{23}>1\)

\(\Rightarrow\dfrac{23}{24}< \dfrac{24}{23}\)

b) \(\dfrac{4}{21}< \dfrac{4}{20}=\dfrac{1}{5}=\dfrac{6}{30}< \dfrac{6}{29}\)

c) \(\dfrac{6}{7}=1-\dfrac{1}{7}< \dfrac{8}{9}=1-\dfrac{1}{9}\)

d) \(\dfrac{1212}{1313}=\dfrac{12\times101}{13\times101}=\dfrac{12}{13}\)

Ta có tỉ thức : \(\dfrac{3}{7} = \dfrac{9}{{21}}\)

Xét \(\dfrac{{3 + 9}}{{7 + 21}}\) = \(\dfrac{{12}}{{28}}\) = \( = \dfrac{3}{7}\)( chia cả tử và mẫu cho 4 )

Xét \(\dfrac{{3 - 9}}{{7 - 21}}\) = \(\dfrac{{ - 6}}{{ - 14}}\)\( = \dfrac{3}{7}\)( chia cả tử và mẫu cho 2 )

Sau khi thực hiện tính các tỉ số ta thấy các kết quả sau khi tối giản của tỉ số bằng với các tỉ só trong tỉ lệ thức đã cho.