Cho minh hoi vai bai:
Bài 1: chứng minh
A=1/100+1/101+...+1/200>1/2
B=1/21+1/22+1/23+... Chung minh 5/6<B<4/3
C=1/2^2+1/3^2+...+1/99^2+1/100^2 <3/4
Bai2: tìm x, y
A) x/3-1/y=1/3
B)1/x+y/2=-1/3
C)x/5-2/y=2/15
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1/5+(1/20+1/21+1/22+1/23+1/24+1/25)+(1/101+1/102+103+104+105) Ta thấy 1/21;1/22;1/23;1/24;1/25 đều nhỏ hơn 1/20 nên 1/21+1/22+1/23+1/24+1/25<5×1/20<1/4 Tương tự 1/101+1/102+1/103+1/104+1/105<5×1/100<1/20 1/5+1/20+1/20=6/20=3/10 1/5+(<1/4)+(<1/20)<1/2 1/2=5/10 3/10<5/10 vậy suy ra điều cần chứng minh
1/5+(1/20+1/21+1/22+1/23+1/24+1/25)+(1/101+1/102+103+104+105)
Ta thấy 1/21;1/22;1/23;1/24;1/25 đều nhỏ hơn 1/20 nên
1/21+1/22+1/23+1/24+1/25<5×1/20<1/4
Tương tự
1/101+1/102+1/103+1/104+1/105<5×1/100<1/20
1/5+1/20+1/20=6/20=3/10
1/5+(<1/4)+(<1/20)<1/2
1/2=5/10
3/10<5/10 vậy suy ra điều cần chứng minh
1.Chưng minh rằng
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
Xét: (1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100) =
(1+1/3+1/5+....+1/99) + (1/2+1/4+1/6+...+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1+1/2+1/3+...+1/50) =
1/51+1/52+1/53+ … + 1/100
Hay:
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
Viết lại:
(1+1/3+1/5+ … +1/199) – (1/2+1/4+1/6+ … +1/200) = 1/101+1/102+ … +1/200
Tương tự như trên ta được:
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1/2+1/4+1/6+...+1/200) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1+1/2+1/3+...+1/100) =
1/101+1/102+ … +1/200
Hay:
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
1 .Chưng minh rằng
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
Xét: (1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100) =
(1+1/3+1/5+....+1/99) + (1/2+1/4+1/6+...+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1/2+1/4+1/6+...+1/100) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/99+1/100) - (1+1/2+1/3+...+1/50) =
1/51+1/52+1/53+ … + 1/100
Hay:
(1+1/3+1/5+....+1/99)-(1/2+1/4+1/6+...+1/100)=1/51+1/52+...+1/100
2.Áp dụng phan 1 để chung minh
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
Viết lại:
(1+1/3+1/5+ … +1/199) – (1/2+1/4+1/6+ … +1/200) = 1/101+1/102+ … +1/200
Tương tự như trên ta được:
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1/2+1/4+1/6+...+1/200) x 2 =
(1+1/2+1/3+1/4+1/5+1/6+....+1/199+1/200) - (1+1/2+1/3+...+1/100) =
1/101+1/102+ … +1/200
Hay:
1-1/2+1/3-1/4+.....-1/200=1/101+1/102+.......+1/200
a,1/51 > 1/100
1/52 > 1/100
1/53 > 1/100
...
1/100=1/100
=>H>1/100 + 1/100 + 1/100 +...+1/100
H>50/100=1/2
1/51<1/50
1/52<1/50
....
1/100<1/50
=>H<1/50+1/50+...+1/50
H<50/50=1
Vay1/2<H<1