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3.13579/34567 = 40737/34567 = 34567+6170/34567
3.13580/34569 = 40740/34569 = 34569+6171/34569
vì : 34567+6170/34567 < 34569+6171/34569
nên: 3.13579/34567 < 3.13580/34569
vậy: 13579/34567 < 13580/34569
a: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)
mà 17^19+1>17^18+1
nên A<B
b: \(2C=\dfrac{2^{2021}-2}{2^{2021}-1}=1-\dfrac{1}{2^{2021}-1}\)
\(2D=\dfrac{2^{2022}-2}{2^{2022}-1}=1-\dfrac{1}{2^{2022}-1}\)
2^2021-1<2^2022-1
=>1/2^2021-1>1/2^2022-1
=>-1/2^2021-1<-1/2^2022-1
=>C<D
B=\(\dfrac{10^9+1}{10^{10}+1}< \dfrac{10^5+1+9}{10^{10}+1+9}=\dfrac{10^9+10}{10^{10}+10}=\dfrac{10.\left(10^8+1\right)}{10\left(10^9+1\right)}\)
= A
Bài 1:
1: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)
mà \(17^{19}+1>17^{18}+1\)
nên 17A>17B
hay A>B
2: \(C=\dfrac{98^{99}+98^{10}+1-98^{10}}{98^{89}+1}=98^{10}+\dfrac{1-98^{10}}{98^{89}+1}\)
\(D=\dfrac{98^{98}+98^{10}+1-98^{10}}{98^{88}+1}=98^{10}+\dfrac{1-98^{10}}{98^{88}+1}\)
mà \(98^{89}+1>98^{88}+1\)
nên C>D
M=\(\dfrac{1919\times171717}{191919\times1717}\) và N=\(\dfrac{18}{19}\)
Ta có :
M= \(\dfrac{1919\times171717}{191919\times1717}\)
M=\(\dfrac{19\times17}{19\times17}\)
M= 1
Mà N= \(\dfrac{18}{19}\)
Vì: 1>\(\dfrac{18}{19}\)
\(\Rightarrow\)\(\dfrac{1919\times171717}{191919\times1717}\) > \(\dfrac{18}{19}\)
\(\Rightarrow\)M > N
A=\(\dfrac{5^{12}+1}{5^{13}+1}\) và B =\(\dfrac{5^{11}+1}{5^{12}+1}\)
Ta có:
A=\(\dfrac{5^{12}+1}{5^{13}+1}\)
\(\Rightarrow\)5.A=5.\(\dfrac{5^{12}+1}{5^{13}+1}\)
=\(\dfrac{5.\left(5^{12}+1\right)}{5^{13}+1}\)
=\(\dfrac{5^{13}+6}{5^{13}+1}\)
=\(\dfrac{\left(5^{13}+1\right)+6}{5^{13}+1}\)
=\(\dfrac{5^{13}+1}{5^{13}+1}\) + \(\dfrac{6}{5^{13}+1}\)
= 1 + \(\dfrac{6}{5^{13}+1}\)
B=\(\dfrac{5^{11}+1}{5^{12}+1}\)
\(\Rightarrow\)5.B = 5.\(\dfrac{5^{11}+1}{5^{12}+1}\)
=\(\dfrac{5.\left(5^{11}+1\right)}{5^{12}+1}\)
=\(\dfrac{5^{12}+6}{5^{12}+1}\)
=\(\dfrac{\left(5^{12}+1\right)+5}{5^{12}+1}\)
=\(\dfrac{5^{12}+1}{5^{12}+1}\) + \(\dfrac{5}{5^{12}+1}\)
= 1 + \(\dfrac{5}{5^{12}+1}\)
Vì: \(5^{13}+1\) > \(5^{12}+1\)
\(\Rightarrow\) \(\dfrac{5}{5^{13}+1}\) < \(\dfrac{5}{5^{12}+1}\)
\(\Rightarrow\) 1+\(\dfrac{5}{5^{13}+1}\) < 1+\(\dfrac{5}{5^{12}+1}\)
\(\Rightarrow\) 5.A < 5.B
\(\Rightarrow\) A < b
Ta có F < 1 nên
\(\dfrac{13580}{34569}< 1< \dfrac{13580+\left(-1\right)}{34569+\left(-1\right)}=\dfrac{13579}{34568}\)<\(\dfrac{13579}{34567}\)
Từ đó suy ra \(\dfrac{13580}{34569}< \dfrac{13579}{34567}\)hay\(\dfrac{13579}{34567}>\dfrac{13580}{34569}\)
Vậy E > F