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a ) − 1212 − 2424 = ( − 1212 ) : ( − 1212 ) ( − 2424 ) : ( − 1212 ) = 1 2
b ) 120120 − 240240 = 120120 : ( − 120120 ) − 240240 : ( − 120120 ) = − 1 2
c) 1313 − 1414 = 1313 : ( − 101 ) − 1414 : ( − 101 ) = − 13 14
a) 33 66 = 33 : 33 66 : 33 = 1 2 b ) − 22 77 = − 22 : 11 77 : 11 = − 2 7
c ) 3030 6060 = 3030 : 3030 6060 : 3030 = − 1 2 d ) − 1212 − 2424 = ( − 1212 ) : ( − 1212 ) ( − 2424 ) : ( − 1212 ) = 1 2
e ) 120120 − 240240 = 120120 : ( − 120120 ) − 240240 : ( − 120120 ) = − 1 2
f ) 1313 − 1414 = 1313 : ( − 101 ) − 1414 : ( − 101 ) = − 13 14
Lúc nãy, cô còn dạy học nên giờ cô mới giảng cho em được nhé.
B = (1 - \(\dfrac{1}{2}\))\(\times\)(1 - \(\dfrac{1}{3}\))\(\times\)(1 - \(\dfrac{1}{4}\))\(\times\)(1-\(\dfrac{1}{5}\))\(\times\)...\(\times\)(1- \(\dfrac{1}{2003}\))\(\times\)(1-\(\dfrac{1}{2004}\))
B = \(\dfrac{2-1}{2}\)\(\times\)\(\dfrac{3-1}{3}\)\(\times\)\(\dfrac{4-1}{4}\)\(\times\)\(\dfrac{5-1}{5}\)\(\times\)...\(\times\)(\(\dfrac{2003-1}{2003}\))\(\times\)(\(\dfrac{2004-1}{2004}\))
B = \(\dfrac{1}{2}\)\(\times\)\(\dfrac{2}{3}\)\(\times\)\(\dfrac{3}{4}\)\(\times\)\(\dfrac{4}{5}\)\(\times\)...\(\times\)\(\dfrac{2002}{2003}\)\(\times\)\(\dfrac{2003}{2004}\)
B = \(\dfrac{2\times3\times4\times...\times2003}{2\times3\times4\times...\times2003}\)\(\times\) \(\dfrac{1}{2004}\)
B = \(\dfrac{1}{2004}\)
a. \(\frac{26}{42}< \frac{56}{78}\)
b.\(\frac{58}{89}< \frac{36}{53}\)
c.\(\frac{1530}{1632}>\frac{1414}{1515}\)
d.\(\frac{-373737}{515151}>\frac{-1111}{1212}\)
k cho mk nha!!!!
\(\dfrac{212121}{232323}=\dfrac{212121:10101}{232323:10101}=\dfrac{21}{23}\)
\(\dfrac{123123}{124124}=\dfrac{123123:1001}{124124:1001}=\dfrac{123}{124}\)
Ta có: \(1-\dfrac{21}{23}=\dfrac{2}{23}\) ; \(1-\dfrac{123}{124}=\dfrac{1}{124}=\dfrac{2}{248}\)
\(=>\dfrac{2}{23}>\dfrac{2}{248}\)
\(=>1-\dfrac{21}{23}>1-\dfrac{123}{124}\)
\(=>\dfrac{21}{23}< \dfrac{123}{124}\)
\(=>\dfrac{212121}{232323}< \dfrac{123123}{124124}\)
Ta có \(A=\frac{20132013}{20142014}=\frac{20132013\div10001}{20142014\div10001}=\frac{2013}{2014}=1-\frac{1}{2014}\)
\(B=\frac{1313}{1414}=\frac{1313\div101}{1414\div101}=\frac{13}{14}=1-\frac{1}{14}\)
Ta thấy \(1=1;\frac{1}{14}>\frac{1}{2014}\Rightarrow1-\frac{1}{14}< 1-\frac{1}{2014}\)
Do đó \(\frac{20132013}{20142014}>\frac{1313}{1414}\)hay \(A>B\)
