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ta có:1/11 ; 1/12 ; 1/13; ....; 1/19 ;1/20 đều lớn hơn 1/20
=>1/11 + 1/12 + 1/13 + ... + 1/19 + 1/20 > 1/20+1/20+...+1/20 (10 phân số 1/20)
=>1/11 + 1/12 + 1/13 + ... + 1/19 + 1/20 > 10/20
=>1/11 + 1/12 + 1/13 + ... + 1/19 + 1/20 > 1/2
=>đpcm
Ta có S = 1/11+1/12+1/13+...+1/19+1/20 nên S có 10 số hạng
Và 1/2 = 10/20 =
Mà 1/11 > 1/12 > 1/13 > 1/14 > 1/15 > 1/16 > 1/17 > 1/18 > 1/19 > 1/20
Nên 1/11+1/12+1/13+...+1/19+1/20 > 1/20x10
=> 1/11+1/12+1/13+...+1/19+1/20 > 10/20
=> 1/11+1/12+1/13+...+1/19+1/20 > 1/2
Vậy S > 1/2Ta có S = 1/11+1/12+1/13+...+1/19+1/20 nên S có 10 số hạng
Và 1/2 = 10/20 =
Mà 1/11 > 1/12 > 1/13 > 1/14 > 1/15 > 1/16 > 1/17 > 1/18 > 1/19 > 1/20
Nên 1/11+1/12+1/13+...+1/19+1/20 > 1/20x10
=> 1/11+1/12+1/13+...+1/19+1/20 > 10/20
=> 1/11+1/12+1/13+...+1/19+1/20 > 1/2
Vậy S > 1/2
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{20}-\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\) (đpcm)
Ta có 1/20 + 1/20 + 1/20 + ... + 1/20 + 1/20 < 1/11 + 1/12 + 1/13 + ... + 1/19 + 1/20 < 1/10 + 1/10 + 1/10 + ... + 1/10 + 1/10 = 10/20 < S < 10/10 \(\Rightarrow\)1/2 < S < 1 ( đpcm )
Ta có : 1/11+1/12+1/13+...+1/19+1/20 > 1/20+1/20+1/20+...+1/20+1/20 =10/20=1/2
có tất cả 10 phân số 1/20
=> S > 1/2
1/11+1/12+1/13+...+1/19+1/20 < 1/10+1/10+1/10+...+1/10+1/10 =10/10=1
có tất cả 10 phân số /10
=> S<1
=> 1/2 < S <1
Ta xét : \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{20}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}\right)\)
\(=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+....+\frac{1}{20}\)
Vì \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+....+\frac{1}{20}=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\)
nên \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+....+\frac{1}{20}\) ( đpcm )
1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...+ 1/19 - 1/20
= ( 1 + 1/3 + 1/5 + ...+ 1/19 ) - ( 1/2 + 1/4 + ...+ 1/20 )
= ( 1 + 1/2 + 1/3 + 1/4 + ...+ 1/19 + 1/20 ) - 2 . ( 1/2 + 1/4 + ...+ 1/20 )
= ( 1 + 1/2 + 1/3 + ...+ 1/20 ) - ( 1 + 1/2 + ... + 1/10 )
= 1/11 + 1/12 + 1/13 + ...+ 1/20 ( Đpcm )
TK mk nha !!!
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{19}-\frac{1}{20}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{19}+\frac{1}{20}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}-1+\frac{1}{2}+....+\frac{1}{10}\)
\(=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\left(đpcm\right)\)
Xét: 1-1/2+1/3-1/4+...+1/19-1/20 = (1+1/3+1/5+...1/19) - (1/2+1/4+1/6+...+1/20)
= (1+ 1/2+1/3+...+1/20) - 2.(1/2+1/4+...+1/20)
= (1+1/2+1/3+...+1/20) - (1+1/2+...+1/10)
= 1/11+1/12+1/13+...+1/20 (dpcm)
Vậy, 1-1.2+1/3-1/4+...+1/19-1/20=1/11+1/12+1/13+...+1/20