Câu 5 : A= 1212 +122122+ 123123+ 124124+ ....+122021122021+122022122022và B= 1313+1414+1515+17601760
a) Rút gọn A
b) So sánh A và B
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a) \(\dfrac{12}{14}=\dfrac{1200}{1400}=\dfrac{1400-200}{1400}=1-\dfrac{200}{1400}\)
\(\dfrac{1212}{1414}=\dfrac{1414-200}{1414}=1-\dfrac{200}{1414}\)
vì \(\dfrac{200}{1414}< \dfrac{200}{1400}\)
Nên \(1-\dfrac{200}{1400}< 1-\dfrac{200}{1414}\)
Vậy \(\dfrac{12}{14}< \dfrac{1212}{1414}\)
Các bài sau tương tự
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
A)1212/1313<1313/1414.B)1717/2525<515151/727272.23/28,7/9,5/7,11/18
\(\dfrac{121212}{131313}=\dfrac{121212:10101}{131313:10101}=\dfrac{12}{13}=1-\dfrac{1}{13}\)
\(\dfrac{1313}{1414}=\dfrac{1313:101}{1414:101}=\dfrac{13}{14}=1-\dfrac{1}{14}\)
Vì \(\dfrac{1}{13}>\dfrac{1}{14}\) nên \(\dfrac{12}{13}< \dfrac{13}{14}\) hay \(\dfrac{1212}{1313}< \dfrac{1313}{1414}\)
b) \(\dfrac{1717}{2525}=\dfrac{1717:101}{2525:101}=\dfrac{17}{25}=\dfrac{51}{75}\)
\(\dfrac{515151}{727272}=\dfrac{515151:10101}{727272:10101}=\dfrac{51}{72}\)
Vì \(\dfrac{51}{75}< \dfrac{51}{72}\) nên \(\dfrac{17}{25}< \dfrac{51}{72}\) hay \(\dfrac{1717}{2525}< \dfrac{515151}{727272}\)
a) \(\dfrac{1212}{1313}=\dfrac{101x12}{101x13}=\dfrac{12}{13}< \dfrac{12+1}{13+1}=\dfrac{13}{14}\)
\(\dfrac{1313}{1414}=\dfrac{101x13}{101x14}=\dfrac{13}{14}\)
Vậy \(\dfrac{1212}{1313}< \dfrac{1313}{1414}\)
Làm tương tự câu b
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 = \(\dfrac{1}{2004}\)