Question

So sánh A=\(\frac{10^{2017}+1}{10^{ }^{2016}+1}\)B=\(\frac{10^{2018}+1}{10^{2017}^{ }+1}\)

Answer 1

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Answer 2

Tường đây

Answer 3

Ta có công thức : 

\(\frac{a}{b}>\frac{a+c}{b+c}\) \(\left(\frac{a}{b}>1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(B=\frac{10^{2018}+1}{10^{2017}+1}>\frac{10^{2018}+1+9}{10^{2018}+1+9}=\frac{10^{2018}+10}{10^{2018}+10}=\frac{10\left(10^{2017}+1\right)}{10\left(10^{2016}+1\right)}=\frac{10^{2017}+1}{10^{2016}+1}=A\)

\(\Rightarrow\)\(B>A\) hay \(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~ 

Answer 4

Có \(A=\frac{10^{2017}+1}{10^{2016}+1}\); \(B=\frac{10^{2018}+1}{10^{2017}+1}\)

Áp dụng công thức \(\frac{a}{b}>\frac{a+c}{b+c}\)( a;b;c \(\inℕ^∗\)):

\(B=\frac{10^{2018}+1}{10^{2017}+1}>\frac{10^{2018}+1+9}{10^{2017}+1+9}=\frac{10^{2018}+10}{10^{2017}+10}=\frac{10\left(10^{2017}+1\right)}{10\left(10^{2016}+1\right)}=\frac{10^{2017}+1}{10^{2016}+1}=A\)

\(\Rightarrow B>A\)hay A < B

Vậy A < B