Question

giúp mik với,chiều thi rùi

So sánh A với B biết

A=\(\dfrac{19^{18}+1}{19^{19}+1}\)

B=\(\dfrac{19^{17}+1}{19^{18}+1}\)

Answer 1

19A= \(\dfrac{19^{19}+19}{19^{19}+1}=\dfrac{19^{19}+1+18}{19^{19}+1}=1+\dfrac{18}{19^{19}+1}\)

19B = \(\dfrac{19^{18}+19}{19^{18}+1}=\dfrac{19^{18}+1+18}{19^{18}+1}=1+\dfrac{18}{19^{18}+1}\)

Ta có: 19A<19B

=> A<B

Answer 2

Ta có:

\(A=\dfrac{19^{18}+1}{19^{19}+1}\)

\(\Rightarrow10A=\dfrac{19^{19}+10}{19^{19}+1}=\dfrac{19^{19}+1+9}{19^{19}+1}=1+\dfrac{9}{19^{19}+1}\)

Ta có:

\(B=\dfrac{19^{17}+1}{19^{18}+1}\)

\(\Rightarrow10B=\dfrac{19^{18}+10}{19^{18}+1}=\dfrac{19^{18}+1+9}{19^{18}+1}=1+\dfrac{9}{19^{18}+1}\)

Vì: 10A<10B

=> A<B

Answer 3

Xin lỗi mình bị nhầm

Answer 4

\(19A=\dfrac{19^{19}+19}{19^{19}+1}=\dfrac{19^{19}+1+18}{19^{19}+1}=1+\dfrac{18}{19^{19}+1}\)

\(19B=\dfrac{19^{18}+19}{19^{18}+1}=\dfrac{19^{18}+1+18}{19^{18}+1}=1+\dfrac{18}{19^{18}+1}\)

Ta có: 10A < 10B

=> A <B