Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta dùng bất đẳng thức\(\frac{a}{b}<\frac{a+n}{b+n}\left(n\ne0\right)\)
Ta có \(B=\frac{10^{20}+1}{10^{21}+1}<\frac{10^{20}+1+9}{10^{21}+1+9}<\frac{10^{20}+10}{10^{21}+10}<\frac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\)
\(<\frac{10^{19}+1}{10^{20}+1}\)
Vậy \(A>B\)
Ta có:\(B=\frac{10^{20}+1}{10^{21}+1}< 1\Rightarrow B=\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)
=> A > B
Ta có :
\(\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}\)
Vậy \(\frac{10^{19}+1}{10^{20}+1}>\frac{10^{20}+1}{10^{21}+1}\)
C2:A=\(\frac{20^{10}+1}{20^{10}-1}\) B=\(\frac{20^{10}-1}{20^{10}-3}\)
A=\(\frac{20^{10}-1+2}{20^{10}-1}\) B=\(\frac{20^{10}-3+2}{20^{10}-3}\)
A=\(\frac{20^{10}-1}{20^{10}-1}\)+\(\frac{2}{20^{10}-1}\) B=\(\frac{20^{10}-3}{20^{10}-3}\)+\(\frac{2}{20^{10}-3}\)
A=1+\(\frac{2}{20^{10}-1}\) B=1+\(\frac{2}{20^{10}-3}\)
Vì 2010-1>2010-3=>\(\frac{2}{20^{10}-1}\)<\(\frac{2}{20^{10}-3}\)
=>1+\(\frac{2}{20^{10}-1}\)<1+\(\frac{2}{20^{10}-3}\)
=>A<B(đây là cách để đi thi còn cách kia làm cho nhanh thôi)
Vì B=\(\frac{20^{10}-1}{20^{10}-3}\)>1
Theo công thức \(\frac{a}{b}\)>1=>\(\frac{a+n}{b+n}\)>\(\frac{a}{b}\)
=>B>\(\frac{20^{10}-1+2}{20^{10}-3+2}\)
=>B>\(\frac{20^{10}+1}{20^{10}-1}\)=A
Vậy B>A
Công thức a/b >1 => a/b > a+n/b+n (a, b,n \(\in\) N*)
B = 2010-1/2010-3 > 1 nên B = 2010-1/2010-3 > 2010-1+2/2010-3+2 = 2010+1/ 2010-1 = A
Vậy A < B
Ta có:
\(20A=\frac{20\left(20^{19}+1\right)}{20^{20}+1}=\frac{20^{20}+20}{20^{20}+1}=\frac{20^{20}+1+19}{20^{20}+1}=\frac{20^{20}+1}{20^{20}+1}+\frac{19}{20^{20}+1}=1+\frac{19}{20^{20}+1}\)
\(20B=\frac{20\left(20^{20}+1\right)}{20^{21}+1}=\frac{20^{21}+20}{20^{21}+1}=\frac{20^{21}+1+19}{20^{21}+1}=\frac{20^{21}+1}{20^{21}+1}+\frac{19}{20^{21}+1}=1+\frac{19}{20^{21}+1}\)
Vì 2020+1<2021+1
\(\Rightarrow\frac{19}{20^{20}+1}>\frac{19}{20^{21}+1}\)
\(\Rightarrow1+\frac{19}{20^{20}+1}>1+\frac{19}{20^{21}+1}\)
\(\Rightarrow20A>20B\)
\(\Rightarrow A>B\)