Ta có: \(20A=\dfrac{20^{19}+20}{20^{19}+1}=1+\dfrac{19}{20^{19}+1}\)
\(20B=\dfrac{20^{18}+20}{20^{18}+1}=1+\dfrac{19}{20^{18}+1}\)
Vì \(\dfrac{19}{20^{19}+1}< \dfrac{19}{20^{18}+1}\Rightarrow1+\dfrac{19}{20^{19}+1}< 1+\dfrac{19}{20^{18}+1}\)
\(\Rightarrow20A< 20B\Rightarrow A< B\)
Vậy A < B
Ta có: \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+c}\)(a \(\in\) N và b,c,d \(\in\) N*)
Áp dụng kiến thức đó, ta được:
A = \(\dfrac{20^{18}+1}{20^{19}+1}\) <\(\dfrac{20^{18}+1+19}{20^{19}+1+19}\)= \(\dfrac{20^{18}+20}{20^{19}+20}\) = \(\dfrac{20\left(20^{17}+1\right)}{20\left(20^{18}+1\right)}\)
= \(\dfrac{20^{17}+1}{20^{18}+1}\) = B
Vậy A < B
Ta có: \(\dfrac{20^{18}+1}{20^{19}+1}< \dfrac{20^{18}+1+19}{20^{19}+1+19}\)
\(\Rightarrow A< \dfrac{20^{18}+20}{20^{19}+20}\)
\(\Rightarrow A< \dfrac{20\left(20^{17}+1\right)}{20\left(20^{18}+1\right)}\)
\(\Rightarrow A< \dfrac{20^{17}+1}{20^{18}+1}\)
\(\Rightarrow A< B\)
Vậy \(A< B\).
Ta có :
A = \(\dfrac{20\cdot\left(20^{18}+1\right)}{20^{19}+1}=\dfrac{20^{19}+20}{20^{19}+1}=\dfrac{\left(20^{19}+1\right)-20}{20^{19}+1}=1-\dfrac{1}{20^{19}+1}\)B = \(\dfrac{20\cdot\left(20^{17}+1\right)}{20^{18}+1}=\dfrac{20^{18}+20}{20^{18}+1}=\dfrac{\left(20^{18}+1.\right)-20}{20^{18}+1}=1-\dfrac{20}{20^{18}+1}\)
Do nên
\(A=\dfrac{20^{18}+1}{20^{19}+1}< \dfrac{20^{18}+1+19}{20^{19}+1+19}\\ =\dfrac{20^{18}+20}{20^{19}+20}=\dfrac{20\left(20^{17}+1\right)}{20\left(20^{18}+1\right)}=\dfrac{20^{17}+1}{20^{18}+1}=B\\ \Rightarrow A< B\)
\(A=\dfrac{20^{18}+1}{20^{19}+1}=\dfrac{20^{18}+1}{20^{19}+1}< 1 =\dfrac{20^{18}+1}{20^{19}+1}< \dfrac{20^{18}+1+19}{20^{19}+1+19}=\dfrac{20^{18}+20}{20^{19}+20}\\ =\dfrac{20\left(20^{17}+1\right)}{20\left(20^{18}+1\right)}=\dfrac{20^{17}+1}{20^{18}+1}=B\)Vậy A < B
