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Ta có : A = \(\frac{10^{2020}+1}{10^{2021}+1}\)
=> 10A = \(\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)
Lại có : \(B=\frac{10^{2021}+1}{10^{2022}+1}\)
=> \(10B=\frac{10^{2022}+10}{10^{2022}+1}=1+\frac{9}{10^{2022}+1}\)
Vì \(\frac{9}{10^{2022}+1}< \frac{9}{10^{2021}+1}\)
=> \(1+\frac{9}{10^{2022}+1}< 1+\frac{9}{10^{2022}+1}\)
=> 10B < 10A
=> B < A
b) Ta có : \(\frac{2019}{2020+2021}< \frac{2019}{2020}\)
Lại có : \(\frac{2020}{2020+2021}< \frac{2020}{2021}\)
=> \(\frac{2019}{2020+2021}+\frac{2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> \(\frac{2019+2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> B < A
a) Ta có A = \(\frac{2^{2018}+1}{2^{2019}+1}\)
=> 2A = \(\frac{2^{2019}+2}{2^{2019}+1}=1+\frac{1}{2^{2019}+1}\)
Lại có B = \(\frac{2^{2017}+1}{2^{2018}+1}\)
=> 2B = \(\frac{2^{2018}+2}{2^{2018}+1}=\frac{2^{2018}+1+1}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)
Vì \(\frac{1}{2^{2018}+1}>\frac{1}{2^{2019}+1}\Rightarrow1+\frac{1}{2^{2018}+1}>1+\frac{1}{2^{2019}+1}\Rightarrow2B>2A\Rightarrow B>A\)
Ta có : A = \(\frac{10^{2020}+1}{10^{2019}+1}\)
=> \(\frac{A}{10}=\frac{10^{2020}+1}{10^{2020}+10}=\frac{10^{2020}+10-9}{10^{2020}+10}=1-\frac{9}{10^{2020}+10}\)
Lại có : B = \(\frac{10^{2021}+1}{10^{2020}+1}\)
=> \(\frac{B}{10}=\frac{10^{2021}+1}{10^{2021}+10}=\frac{10^{2021}+10-9}{10^{2021}+10}=1-\frac{9}{10^{2021}+10}\)
Vì : \(\frac{9}{10^{2021}+10}< \frac{9}{10^{2020}+10}\Rightarrow1-\frac{9}{10^{2021}+10}>1-\frac{9}{10^{2020}+10}\Rightarrow\frac{B}{10}>\frac{A}{10}\Rightarrow B>A\)
Vậy B > A
a, Ta có : \(10^{15}\cdot11=10^{15}\left(10+1\right)=10^{16}+10^{15}\)
Vì \(10^{16}+10^{15}>10^{16}+10\)
\(\Rightarrow\dfrac{10^{16}+10^{15}}{10^{16}+1}>\dfrac{10^{16}+10}{10^{16}+1}\)
Hay A>B
b, Ta có : \(C=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}}{10^{10}-1}+\dfrac{1}{10^{10}-1}\)
\(D=\dfrac{10^{10}-1}{10^{13}-3}=\dfrac{10^{10}}{10^{13}-3}+\dfrac{-1}{10^{13}-3}\)
Vì \(\dfrac{10^{10}}{10^{10}-1}>\dfrac{10^{10}}{10^{13}-3};\dfrac{1}{10^{10}-1}>\dfrac{-1}{10^{13}-3}\)
\(\Rightarrow\dfrac{10^{10}+1}{10^{10}-1}>\dfrac{10^{10}-1}{10^{13}-3}\)
Hay C > D
a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
Vậy A < B
b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)
\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)
Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)
c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:
\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
Vậy A < B
D= 1/2. (1/25-1/27 +1/27-1/29+...+1/73-1/75)
= 1/2. (1/25 -1/75)
=1/2 . 2/75= 1/75
D = \(\dfrac{1}{25.27}+\dfrac{1}{27.29}+...+\dfrac{1}{73.75}\)
2D = 2( \(\dfrac{1}{25.27}+\dfrac{1}{27.29}+...+\dfrac{1}{73.75}\) )
= \(\dfrac{2}{25.27}+\dfrac{2}{27.29}+...+\dfrac{2}{73.75}\)
= \(\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\)
= \(\dfrac{1}{25}-\dfrac{1}{75}\)
= \(\dfrac{2}{75}\)
d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
\(n\left(n+3\right)=n^2+3n\)
\(\left(n+2\right)\left(n+1\right)=n^2+3n+2\)
Vì \(n^2+3n< n^2+3n+2\Rightarrow\dfrac{n}{n+1}< \dfrac{n+2}{n+3}\left(n\in N\right)\)
b) \(\dfrac{n}{2n+1}=\dfrac{3n}{6n+3}< \dfrac{3n+1}{6n+3}\)
c) \(\dfrac{10^8+2}{10^8-1}=1+\dfrac{1}{10^8-1}\)
\(\dfrac{10^8}{10^8-3}=\left(1+\dfrac{3}{10^8-3}\right)\)
Vì \(\dfrac{1}{10^8-1}>\dfrac{3}{10^8-3}\Rightarrow\dfrac{10^8+2}{10^8-1}< \dfrac{10^8}{10^8-3}\)
Làm dần dần và làm từ từ, suy ra được nhiều cách giải.
a) \(\dfrac{n}{n+1}\) và \(\dfrac{n+2}{n+3}\)
+ Cách 1:
\(\dfrac{n}{n+1}=\dfrac{n+1-1}{n+1}=1-\dfrac{1}{n+1}\)
\(\dfrac{n+2}{n+3}=\dfrac{n+3-1}{n+3}=1-\dfrac{1}{n+3}\)
Vì \(\dfrac{1}{n+1}>\dfrac{1}{n+3}\) nên \(1-\dfrac{n}{n+1}< 1-\dfrac{1}{n+3}\)
\(\Rightarrow\dfrac{n}{n+1}< \dfrac{n+2}{n+3}\)
+ Cách 2:
Ta so sánh: \(n\left(n+3\right)\) và \(\left(n+1\right)\left(n+2\right)\)
\(n\left(n+3\right)=nn+3n=n^2+3n\)
\(\left(n+1\right)\left(n+2\right)=\left(n+1\right)n+\left(n+1\right).2=n^2+n+2n+2=n^2+3n+2\)
Vì \(n^2+3n< n^2+3n+2\) nên \(\dfrac{n}{n+1}< \dfrac{n+2}{n+3}\)
b) \(\dfrac{n}{2n+1}\) và \(\dfrac{3n+1}{6n+3}\)
Ta so sánh: \(n\left(6n+3\right)\) và \(\left(2n+1\right)\left(3n+1\right)\)
\(n\left(6n+3\right)=n.6n+3n=6n^2+3n\)
\(\left(2n+1\right)\left(3n+1\right)=\left(2n+1\right)3n+\left(2n+1\right)=6n^2+3n+2n+1=6n^2+5n+1\)
Vì \(6n^2+3n< 6n^2+5n+1\) nên \(\dfrac{n}{2n+1}< \dfrac{3n+1}{6n+3}\)
c) \(\dfrac{10^8+2}{10^8-1}\) và \(\dfrac{10^8}{10^8-3}\)
\(\dfrac{10^8+2}{10^8-1}=\dfrac{10^8-1+3}{10^8-1}=1+\dfrac{3}{10^8-1}\)
\(\dfrac{10^8}{10^8-3}=\dfrac{10^8-3+3}{10^8-3}=1+\dfrac{3}{10^8-3}\)
Vì \(\dfrac{3}{10^8-1}>\dfrac{3}{10^8-3}\) nên \(\dfrac{10^8+2}{10^8-1}>\dfrac{10^8}{10^8-3}\)
d) \(\dfrac{3^{17}+1}{3^{20}+1}\) và \(\dfrac{3^{20}+1}{3^{23}+1}\)
(đang tìm cách làm, và thêm vài cách khác)
b: \(A=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)
\(B=\dfrac{10^8-7+13}{10^8-7}=1+\dfrac{13}{10^8-7}\)
mà \(10^7-8< 10^8-7\)
nên A>B
c: \(\dfrac{1}{10}A=\dfrac{10^{1992}+1}{10^{1992}+10}=1-\dfrac{9}{10^{1992}+10}\)
\(\dfrac{1}{10}B=\dfrac{10^{1993}+1}{10^{1993}+10}=1-\dfrac{9}{10^{1993}+10}\)
mà \(\dfrac{9}{10^{1992}+10}>\dfrac{9}{10^{1993}+10}\)
nên A<B
\(C=\dfrac{10^{2021}+10-9}{10^{2020}+1}=10-\dfrac{9}{10^{2020}+1}\)
\(D=\dfrac{10^{2022}+10-9}{10^{2021}+1}=10-\dfrac{9}{10^{2021}+1}\)
mà \(10^{2020}+1< 10^{2021}+1\)
nên \(-\dfrac{9}{10^{2020}+1}< -\dfrac{9}{10^{2021}+1}\)
hay C<D