So Sánh
a= 112/111+113/112+114/113
b= 111/112+112/111
tính bằng cách thuận tiện
111/110 * 112/111 * 113/112 * 114/113 * 55/57
giúp mk nha mk đg cần gấp thạnk you
\(=\dfrac{1}{110}\times\dfrac{1}{1}\times\dfrac{1}{1}\times\dfrac{114}{1}\times\dfrac{55}{57}\\ =\dfrac{114\times55}{110\times57}\\ =\dfrac{2\times1}{2\times1}=\dfrac{2}{2}=1\)
\(\dfrac{111}{110}\times\dfrac{112}{111}\times\dfrac{113}{112}\times\dfrac{114}{113}\times\dfrac{55}{57}=\dfrac{111\times112\times113\times114}{110\times111\times112\times113}\times\dfrac{55}{57}\)
\(=\dfrac{114}{110}\times\dfrac{55}{57}=\dfrac{57}{55}\times\dfrac{55}{57}=1\)
Cho:
S = 1 11 + 1 12 + 1 13 + 1 14 + 1 15 + 1 16 + 1 17 + 1 18 + 1 19 + 1 20
Hãy so sánh S và 1 2
Chứng minh rằng:
a ) A = 1 12 + 1 13 + 1 14 + ... + 1 22 > 1 2 b ) B = 1 6 + 1 7 + 1 8 + ... + 1 18 + 1 19 < 2 c ) C = 1 10 + 1 11 + 1 12 + ... + 1 99 + 1 100 > 1
a) A = 1 12 + 1 13 + 1 14 + ... + 1 22 > 1 22 + 1 22 + ... 1 22 ⏟ 11 s = 11 22 = 1 2 .
b) B = 1 6 + ... 1 9 + 1 10 + ... + 1 19 < 1 4 + ... + 1 4 ⏟ 4 s o + 1 10 + ... + 1 10 ⏟ 10 s o = 2
c) C = 1 10 + 1 11 + ... + 1 100 > 1 10 + 1 100 = ... + 1 100 ⏟ 90 s o = 1 10 + 90 100 = 1
119 + 118 + 117 + 116 + 115 + 114 + 113 + 112 + 111 + 100 = giúp mik nha
= 1135 nha ban chuc ban hoc gioi
C = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + . . . - 111 - 112 + 113 + 114 - 115
C = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... - 111 - 112 + 113 + 114 - 115 ( có 115 số; 115 chia 4 dư 2)
C = 1 + (2 - 3 - 4 + 5) + (6 - 7 - 8 + 9) + ... + (110 - 111 - 112 + 113) + 114 - 115
C = 1 + 0 + 0 + ... + 0 + 114 - 115
C = 1 + 114 - 115
C = 115 - 115
C = 0
Chứng minh rằng:
a ) A = 1 11 + 1 12 + 1 13 + ... + 1 20 > 1 2 b ) B = 1 5 + 1 6 + 1 7 + ... + 1 16 + 1 17 < 2 c ) C = 1 10 + 1 11 + 1 12 + ... + 1 18 + 1 19 < 1
a) A > 1 20 + 1 20 + ... + 1 20 ⏟ 10 s o = 10 20 = 1 2 .
b) B = 1 5 + ... 1 9 + 1 10 + ... + 1 17 < 1 5 + ... + 1 5 ⏟ 5s o + 1 8 + ... + 1 8 ⏟ 8s o = 2
c) C = 1 10 + 1 11 + 1 12 ... + 1 18 + 1 19 < 1 10 + 1 10 + ... 1 10 ⏟ 9 s o = 1
Cho A=50/111+50/112+50/113+50/114.Chứng tỏ 1<A<2
Ta có :
\(A=\dfrac{50}{111}+\dfrac{50}{112}+\dfrac{50}{113}+\dfrac{50}{114}\)
Ta thấy :
\(\dfrac{50}{111}>\dfrac{50}{200}\)
\(\dfrac{50}{112}>\dfrac{50}{200}\)
\(\dfrac{50}{113}>\dfrac{50}{200}\)
\(\dfrac{50}{114}>\dfrac{50}{200}\)
\(\Rightarrow A>\dfrac{50}{200}+\dfrac{50}{200}+\dfrac{50}{200}+\dfrac{50}{200}\)
\(\Rightarrow A>\dfrac{50}{200}.4=1\) \(\left(1\right)\)
Mặt khác :
\(\dfrac{50}{111}< \dfrac{50}{100}\)
\(\dfrac{50}{112}< \dfrac{50}{100}\)
\(\dfrac{50}{113}< \dfrac{50}{100}\)
\(\dfrac{50}{114}< \dfrac{50}{100}\)
\(\Rightarrow A< \dfrac{50}{100}+\dfrac{50}{100}+\dfrac{50}{100}+\dfrac{50}{100}\)
\(\Rightarrow A< \dfrac{50}{100}.4=2\) \(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Rightarrow1< A< 2\rightarrowđpcm\)
Tính tổng sau G= 1+2-3-4+5+6-7-8+...-111-112+113+114+115
Tính :
C = 1 + 2 - 3 - 4 + 5 + ... - 111 - 112 + 113 + 114 + 115
C = 1 + 2 - 3 - 4 + 5 + ... - 111 - 112 + 113 + 114 + 115 ( có 115 số; 115 : 4 dư 3)
C = 1 + (2 - 3 - 4 + 5) + (6 - 7 - 8 + 9) + ... + (110 - 111 - 112 + 113) + 114 + 115
C = 1 + 0 + 0 + ... + 0 + 114 + 115
C = 1 + 114 + 115
C = 115 × 2
C = 230