Cho A =+++....+.C/minh A <
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(1+2+3+4+...+2014)*(127*126126-127127*126)
=(1+2+3+4+...+2014)*(127*1001*126-127127*126)
=(1+2+3+4+...+2014)*(127127*126-127127*126)
=(1+2+3+4+...+2014)*0
=0
l-i-k-e cho mình nha bạn.
( 1 + 2 + 3 + 4 + 5 + ... + 2018 ) x ( 126 x 127127 - 127127 x 126 )
= ( 1 + 2 + 3 + 4 + 5 + ... + 2018 ) x [( 126 - 126 ) x 127127 ]
= ( 1 + 2 + 3 + 4 + 5 + ... + 2018 ) x ( 0 x 127127 )
= ( 1 + 2 + 3 + 4 + 5 + ... + 2018 ) x 0
= 0
Tk nhé !
#tksnhieu
( 1 + 2 + 3 + 4 + 5 + ... + 2018 ) x ( 126 x 127127 - 127127 x 126 )
= ( 1 + 2 + 3 + 4 + 5 + ... + 2018 ) x [( 126 - 126 ) x 127127 ]
= ( 1 + 2 + 3 + 4 + 5 + ... + 2018 ) x ( 0 x 127127 )
= ( 1 + 2 + 3 + 4 + 5 + ... + 2018 ) x 0
= 0
a, 12/19 và 22/33
ta có :
12/19=12/9=12*3/9*3=36/27
22/33=2/3=2*9/3*9=18/27
Và 36/27 > 18/27 => 12/19 > 2/3
Hay 12/19 > 22/33
b, 12/13 và 1212/1313
ta có :
1212/1313=12*101/13*101=12/13
12/13=12/13
Vì 12/13=12/13=> 12/13=1212/1313
\(\dfrac{212121}{232323}=\dfrac{212121:10101}{232323:10101}=\dfrac{21}{23}\)
\(\dfrac{123123}{124124}=\dfrac{123123:1001}{124124:1001}=\dfrac{123}{124}\)
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 124124/125125 = 124/125
1 - 123/124 = 1/124 ; 1 - 124/125 = 1/125
Vì 1/124 > 1/125 nên 124124/125125 > 123/124
Ta có : \(\frac{124124}{125125}=\frac{124124:1001}{125125:1001}=\frac{124}{125}\)
\(\frac{123}{124}=1-\frac{1}{124}\); \(\frac{124}{125}=1-\frac{1}{125}\)
Vì 1/124 < 1/125 => 123/124 > 123/125 => 123/124 > 124124/125125
ta có: \(\frac{124124}{125125}=\frac{124}{125}\)
Lại có: \(1-\frac{123}{124}=\frac{1}{124};1-\frac{124}{125}=\frac{1}{125}\)
\(\Rightarrow\frac{1}{124}>\frac{1}{125}\Rightarrow1-\frac{123}{124}>1-\frac{124}{125}\)
\(\Rightarrow\frac{123}{124}< \frac{124}{125}\Rightarrow\frac{123}{124}< \frac{124124}{125125}\)