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

a, 13/19

b, 5

c, 31/48

d, 6/7

mình cần cách giải

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 !!! 

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{19}-\frac{1}{20}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{19}+\frac{1}{20}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}-1+\frac{1}{2}+....+\frac{1}{10}\)

\(=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\left(đpcm\right)\)

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{19}-\frac{1}{20}\)

= \(\left(1+\frac{1}{3}+\frac{1}{5}+.........+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.........+\frac{1}{20}\right)\)

= \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{19}+\frac{1}{20}-2\left(\frac{1}{2}+\frac{1}{4}+.......+\frac{1}{20}\right)\)

= \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..........+\frac{1}{19}+\frac{1}{20}+1+\frac{1}{2}+.............+\frac{1}{20}\)

= \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+.........+\frac{1}{20}\)

Vậy biểu thức \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.........+\frac{1}{19}-\frac{1}{20}=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+.......+\frac{1}{20}\)( đpcm)

Ta có:\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+.........+\frac{1}{19}+\frac{1}{20}\)

\(>\frac{1}{20}+\frac{1}{20}+........+\frac{1}{20}\)     (có 10 số \(\frac{1}{20}\))

\(=\frac{1}{20}.10=\frac{1}{2}\)

\(\Rightarrow S>\frac{1}{2}\left(đpcm\right)\)

Ta có : 1/11 < 1/20 , 1/12 < 1/20 , .. , 1/19 < 1/20 , 1/20 = 1/20

=> 1/11 + 1/12 + ...+ 1/19 + 1/20 > 1/20 . 10 

=>               S > 10/20

=>               S > 1/2 

Chúc học giỏi !!! ^_^ 

We have S = 1/11 + 1/12 + 1/13 + ... + 1/19 + 1/20 so S has 10 terms 
And 1/2 = 10/20 = 
1/11> 1/12 > 1/13> 1/14> 1/15> 1/16> 1/17> 1/18> 1/19> 1/20 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 / 
S + / + S + 1/2

I love you ^ - ^

$thanks$

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1/2 chứ bn ?