Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{10^{17}+3}{10^{17}+1}=1+\dfrac{2}{10^{17}+1}\\ B=\dfrac{10^{18}+1}{10^{18}-1}=1+\dfrac{2}{10^{18}-1}=1+\dfrac{2}{10^{17}+1+\left(9\cdot10^{17}-2\right)}\)
Ta có : \(9\cdot10^{17}-2>0\Rightarrow10^{17}+1+\left(9\cdot10^{17}-2\right)>10^{17}+1\\ \Rightarrow\dfrac{2}{10^{17}+1}>\dfrac{2}{10^{18}-1}\Rightarrow A>B\)
Ta có :
\(A=\frac{10^{15}+1}{10^{16}+1}=\frac{\left(10^{15}+1\right).10}{\left(10^{16}+1\right).10}=\frac{10^{16}+10}{10^{17}+10}\)
\(\Rightarrow A=\frac{10^{16}+1+9}{10^{17}+1+9}\)
Vì \(\frac{10^{16}+1}{10^{17}+1}< \frac{10^{16}+1+9}{10^{17}+1+9}\)
Mà \(A=\frac{10^{16}+1+9}{10^{17}+1+9}\)nên \(A>B\)
Vậy \(A>B\)
Vì \(20^{10}-1>20^{10}-3\)
\(\Rightarrow B=\frac{20^{10}-1}{20^{10}-3}>1\)
\(\Rightarrow B=\frac{20^{10}-1}{20^{10}-3}>\frac{20^{10}-1+2}{20^{10}-3+2}=\frac{20^{10}+1}{20^{10}-1}=A\)
\(\Rightarrow B>A\)
\(\Rightarrow A< B\)
vậy A < B
Trả lời:
a, Ta có: 320 ; 274 = ( 33 )4 = 312
Vì 320 > 312 nên 320 > 274
b, 225 ; 166 = ( 24 )6 = 224
Vì 225 > 224 nên 225 > 166
a) ta có: 3100 = (32)50 = 950
b) ta có: 330 = (33)10 = 2710 > 810
c) ta có: 36.67 = 62.67 = 69
Lại có: 433 > 427 = (43)9 = 649 > 69
=> 433>36.67
\(a,\)\(3^{100}\)\(=3^{2.50}\)=\(\left(3^2\right)\)\(^{50}\)\(=9^{50}\)
\(\Rightarrow\)\(3^{100}\)= \(9^{50}\)
\(\frac{10^{20}+1}{10^{22}+1}=\frac{10^{20}+\frac{1}{100}+\frac{99}{100}}{10^{22}+1}=\frac{1}{100}+\frac{99}{100\left(10^{22}+1\right)}\)
\(\frac{10^{22}+1}{10^{24}+1}=\frac{10^{22}+\frac{1}{100}+\frac{99}{100}}{10^{24}+1}=\frac{1}{100}+\frac{99}{100\left(10^{24}+1\right)}\)
Có \(10^{22}+1< 10^{24}+1\Rightarrow\frac{99}{100\left(10^{22}+1\right)}>\frac{99}{100\left(10^{24}+1\right)}\)
do đó \(\frac{10^{20}+1}{10^{22}+1}>\frac{10^{22}+1}{10^{24}+1}\).
1.a)A = (1 - 1/3)(1-2/5)...(1-5/5)....(1-9/5)
=(1-1/3)....0.....(1-9/5)
=0
=>đpcm.
b)ta xét:
1/22 = 1/2x2 < 1/1x2
.............
1/82 = 1/8x8 <1/7x8
=>B < 1/1x2 + 1/2x3 ... + 1 + 1/7x8
<=> B <1 - 1/2 + 1/2 - 1/3 + ... + 1/7 - 1/8
<=> B < 1 - 1/8 = 7/8 < 1
=> B < 1 => đpcm
2.a) Đặt m = 2007(2006+2007) = 2006(2006 + 2007) + (2006+2007)
Đặt n = 2006(2007+2008) = 2006(2006+2007) + (2006 + 2006)
Ta thấy : (2006+2007) > (2006 + 2006) => m > n , áp dụng công thức "a.d > c.d <=> a/b > b/d (a,c thuộc Z// b,d thuộc N)
=> A > B
b)ta có: D = 196 + 197/197 + 198 = (196/197+198) + (197/197+198) < 196/197 + 197/198 = C
=> C > D
c)gọi 2010 là a
ta thấy : (a + 1)(a-3) = (a - 1)(a - 3) + 2(a - 3) < (a - 1)(a - 3) + 2(a - 1) = (a - 1)(a - 1)
áp dụng: ad > bc <=> a/b > c/d ( a,b,c,d thuộc Z// b,d > 0)
=> E > F
a
nAK.DNX. 0pwi9dOjkciopjopoijasd