Question

1. Cho C = \(\frac{19^{208}+1}{19^{209}+1}\) ; D = \(\frac{19^{209}+1}{19^{210}+1}\). So sánh C và D.

2. Cho A = \(\frac{2004}{2005}+\frac{2005}{2006}\) và B = \(\frac{2004+2005}{2005+2006}\)

Trong hai số A và B, số nào lớn hơn?

Answer 1

Bài 1:19.C=\(\frac{19^{209}+19}{19^{209}+1}\)=\(\frac{19^{209}+1+18}{19^{209}+1}\)=\(\frac{19^{209}+1}{19^{209}+1}\)+\(\frac{18}{19^{209}+1}\)=1+\(\frac{18}{19^{209}+1}\)19D=\(\frac{19^{210}+19}{19^{210}+1}\)=\(\frac{19^{210}+1+18}{19^{210}+1}\)=\(\frac{19^{210}+1}{19^{210}+1}\)+\(\frac{18}{19^{210}+1}\)=1+\(\frac{18}{19^{210}+1}\).Vì \(\frac{18}{19^{209}+1}\)>\(\frac{18}{19^{210}+1}\)nên 19A>19B\(\Rightarrow\)A>B

Answer 2

19D=\(\frac{\left(19^{209}+1\right).19}{19^{210}+1}=\frac{19^{210}+19}{19^{210}+1}=\frac{\left(19^{210}+1\right)+18}{19^{210}+1}=\frac{19^{210}+1}{19^{210}+1}+\frac{18}{19^{210}+1}=1+\frac{18}{19^{210}+1}\)

Vì 19C>19D nên C>D

Answer 3

Bài 2:Ta có:A=\(\frac{2004}{2005}\)+\(\frac{2005}{2006}\)>\(\frac{2004}{2005+2006}\)+\(\frac{2005}{2005+2006}\)=\(\frac{2004+2005}{2005+2006}\)\(\Rightarrow\)A>B

Answer 4

19C=\(\frac{\left(19^{208}+1\right).19}{\left(19^{209}+1\right).19}=\frac{19^{209}+19}{19^{209}+1}=\frac{19^{209}+1+18}{19^{209}+1}=\frac{\left(19^{209}+1\right)}{19^{209}+1}=\frac{19^{209}+1}{19^{209}+1}+\frac{19}{19^{209}+1}=1+\frac{19}{19^{209}+1}\)

Answer 5

19C=\(\frac{\left(19^{208}+1\right).19}{19^{209}+1}=\frac{19^{209}+19}{19^{209}+1}=\frac{\left(19^{209}+1\right)+18}{19^{209}+1}=\frac{19^{209}+1}{19^{209}+1}+\frac{18}{19^{209}+1}=1+\frac{18}{19^{209}+1}\)

câu đầu mk làm sai nhé .

câu trả lời thứ 2,3 là 1 bài