Question

giúp tôi so sánh bài này:A=^{_{​\(\frac{10^{1992}+1}{10^{1991}+1}\)}}; B=\(\frac{10^{1993}+1}{10^{1992}+1}\)

hãy giúp tôi giải nha!

Answer 1

Ta thấy \(10^{1993}+1>10^{1992}+1\)

\(\Rightarrow B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10.\left(10^{1992}+1\right)}{10.\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)

\(\Rightarrow A< B\)

Answer 2

\(\frac{10^{1993}+1}{10^{1992}+1}>1\)

\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}\)

\(\frac{10^{1993+1}}{10^{1992}+1}>\frac{10^{1993}+10}{10^{1992}+10}\)

\(\Rightarrow\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1992}+1}{10^{1991}+1}\)

Answer 3

thank you.cảm ơn bạn nhưng mình thấy bài giảng này không chắc lắm!

Answer 4

Ta có 10A=