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ta có: \(\frac{73}{75}>\frac{73}{79}>\frac{77}{79}\Rightarrow\frac{73}{75}>\frac{77}{79}\)
ta có: \(\frac{53}{100}< \frac{47}{100}\)
ta có: \(\frac{48}{47}>1;\frac{84}{85}< 1\Rightarrow\frac{48}{47}>\frac{84}{85}\)
TL:
\(\frac{12}{100}\)= 0,12
\(\frac{5}{100}\)= 0,05
\(\frac{306}{1000}\)= 0,306
-HT-
\(\dfrac{47}{95}\) 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 \(\dfrac{783}{901}>\dfrac{738}{915}\)
=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)
a) 78/79 < 1 < 79/78
b) 1 - 135/136 = 1/136
1 - 136/137 = 1/137
mà 1/136 > 1/137
=>135/136 > 136/137
nhớ vho mik nha
a)
\(\frac{78}{79}\)<1<\(\frac{79}{78}\) suy ra 78/ 79 < 79/78
mk chỉ lm đc phần a thôi, cứ k cho mk nhé!!
Ta có công thức tổng quát:
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
\(a,A=\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\\ =\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}\\ =\dfrac{x-2}{15\left(x+3\right)}\)
Theo đề bài ta có:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
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 \(\dfrac{48}{47}>\dfrac{84}{85}\)
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