So sánh A = \(\frac{10^{19}+1}{10^{20}+1}\) và B = \(\frac{10^{20}+1}{10^{21}+1}\) ?
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ND
1
8 tháng 11 2017
\(B=\frac{10^{20}+1}{10^{21}+1}< 1\)
NÊN \(\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)
VẬY B<A
PT
1
24 tháng 10 2016
Ta thấy:A=\(\frac{10^{19}+1}{10^{20}+1}\)=>10A=\(\frac{10^{20}+10}{10^{20}+1}\)
=>10A=\(\frac{10^{20}+1+9}{10^{20}+1}\)
=>10A=1+\(\frac{9}{10^{20}+1}\)
Ta thấy:B=\(\frac{10^{20}+1}{10^{21}+1}\)
=>10B=\(\frac{10^{21}+10}{10^{21}+1}\)
=>10B=\(\frac{10^{21}+1+9}{10^{21}+1}\)
=>10B=1+\(\frac{9}{10^{21}+1}\)
Do \(\frac{9}{10^{20}+1}\)> \(\frac{9}{10^{21}+1}\)=>A > B
O
1
LN
7 tháng 3 2017
10A=\(\frac{10^{20}+10}{10^{20}+1}\)=\(\frac{10^{20}+1+9}{10^{20}+1}\)=\(1\)+\(\frac{9}{10^{20}+1}\)
10B=\(\frac{10^{21}+10}{10^{21}+1}\)=\(\frac{10^{21}+1+9}{10^{21}+1}\)=\(1\)+\(\frac{9}{10^{21}+1}\)
Vì \(\frac{9}{10^{20}+1}\)>\(\frac{9}{10^{21}+1}\)nên 10A>10B\(\Rightarrow\)A>B
BV
2
20 tháng 7 2018
Ta chứng minh bài toán phụ:
Với a<b thì\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(c\inℕ^∗\right)\)
Ta có: \(a< b\)
\(\Rightarrow ac< bc\)
\(\Rightarrow ac+ba< bc+ba\)
\(a\left(b+c\right)< b.\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\)
đpcm
Áp dụng vào bài toán 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 \(\frac{10^{19}+1}{10^{20}+1}>\frac{10^{20}+1}{10^{21}+1}\)
Tham khảo nhé~
20 tháng 7 2018
Đặt \(A=\frac{10^{19}+1}{10^{20}+1}\)
\(\Rightarrow10A=\frac{10^{20}+10}{10^{20}+1}=\frac{10^{20}+1+9}{10^{20}+1}=1+\frac{9}{10^{20}+1}\)
\(B=\frac{10^{20}+1}{10^{21}+1}\)
\(\Rightarrow10B=\frac{10^{21}+10}{10^{21}+1}=\frac{10^{21}+1+9}{10^{21}+1}=1+\frac{9}{10^{21}+1}\)
\(\Rightarrow\frac{9}{10^{20}+1}>\frac{9}{10^{21}+1}\)
\(\Rightarrow1+\frac{9}{10^{20}+1}>1+\frac{9}{10^{21}+1}\)
\(\Rightarrow10A>10B\Rightarrow A>B\)
27 tháng 4 2016
Do \(B=\frac{10^{20}+1}{10^{21}+1}\)<1
\(\Rightarrow B=\frac{10^{20}+1}{10^{21}+1}\)<\(\frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)
\(\Rightarrow\)B<A hay A<B
22 tháng 2 2018
a) Ta có : B = \(\frac{9^{19}+1}{9^{20}+1}\)< \(\frac{9^{19}+1+8}{9^{20}+1+8}\)= \(\frac{9^{19}+9}{9^{20}+9}\)= \(\frac{9\left(9^{18}+1\right)}{9\left(9^{19}+1\right)}\)= \(\frac{9^{18}+1}{9^{19}+1}\)= A
Vậy A > B
b) Ta có : B = \(\frac{10^{2018}-1}{10^{2019}-1}\)> \(\frac{10^{2018}-1-9}{10^{2019}-1-9}\)= \(\frac{10^{2018}-10}{10^{2019}-10}\)= \(\frac{10\left(10^{2017}-1\right)}{10\left(10^{2018}-1\right)}\)= \(\frac{10^{2017}-1}{10^{2018}-1}\)= A
Vậy A < B.
NHỚ K CHO MK VỚI NHÉ !!!!!!!!
10 tháng 7 2017
B2 :P Ta có : \(B=\frac{2^{10}+1}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(C=\frac{2^{10}-1}{2^{10}-3}=1+\frac{2}{2^{10}-3}\)
Nên : B > C
Áp dụng \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\) (a;b;c \(\in\) N*)
Ta có:
\(B=\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}\)
\(B< \frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)
=> A > B
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