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\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}=\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+...++\frac{1}{299}+\frac{1}{300}\right)\)
\(=\left(\frac{1}{200}.100\right)+\left(\frac{1}{300}.100\right)\)
\(=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}>\frac{4}{6}=\frac{2}{3}\)
\(Vậy\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\RightarrowĐPCM\)
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a,A=1 + ( -3) + 5 + ( -7 ) + ... + 17 + ( -19 )
A=( 1 - 3 ) + ( 5 - 7 ) + ...+ ( 17 +19 )
A= (-2 ) . 10
A= (-20)
b, B= 1-4+7-10 +... -100 + 103
B= 1+ ( -4 + 7 ) + ( -10 +13 ) +...+ (-100 +103 )
B= 1 + 3 + 3 +...+3
B= 1+3 .17
B= 52
c, C= 1 + 2 -3 -4+5+6-7-8+..-99-100+101+102
C= 1 + ( 2-3-4+5) +(6-7 -8+9)+...+(98-99-100+101)+102
C= 1 + 0 + 0 + 0 + 0 + ... + 0 + 102
C= 103
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A < 1/(1.2) + 1/(2.3) + 1/(3.4) + ...+ 1/(99.100)
<=> A< 1- 1/2 + 1/2 - 1/3 + 1/4 - 1/5 + .. + 1/99 - 1/100
<=> A < 1 - 1/100 < 1 (đpcm)
So với thì đây