10a=10^2017+10/10^2017+1
10b=10^2018+10/10^2018+1

cậu tự so sánh nhé vậy là dễ rồi

Ta có: \(A=\dfrac{10^{2016}+1}{10^{2017}+1}\Rightarrow10A=\dfrac{10\left(10^{2016}+1\right)}{10^{2017}+1}=\dfrac{10^{2017}+10}{10^{2017}+1}\)

\(=\dfrac{10^{2017}+1+9}{10^{2017}+1}=\dfrac{10^{2017}+1}{10^{2017}+1}+\dfrac{9}{10^{2017}+1}=1+\dfrac{9}{10^{2017}+1}\)

Tương tự ta cũng có: \(10B=1+\dfrac{9}{10^{2018}+1}\)

Lại có: \(10^{2017}< 10^{2018}\Rightarrow10^{2017}+1< 10^{2018}+1\)

\(\Rightarrow\dfrac{1}{10^{2017}+1}>\dfrac{1}{10^{2018}+1}\Rightarrow\dfrac{9}{10^{2017}+1}>\dfrac{9}{10^{2018}+1}\)

\(\Rightarrow1+\dfrac{9}{10^{2017}+1}>1+\dfrac{9}{10^{2018}+1}\Rightarrow10A>10B\Rightarrow A>B\)

A>B do A>4 cònB<4

Con ngu

Ta có : \(10.A=\frac{10^{2017}+10}{10^{2017}+1}=\frac{10^{2017}+1+9}{10^{2017}+1}=\frac{10^{2017}+1}{10^{2017}+1}+\frac{9}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)

\(10.B=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}+1}=\frac{10^{2018}+1}{10^{2018}+1}+\frac{9}{10^{2018}+1}=1+\frac{9}{10^{2018}+1}\)

Vì \(1=1\)và \(\frac{9}{10^{2017}+1}>\frac{9}{10^{2018}+1}\)nên \(1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)hay \(A>B\)

Vậy \(A>B\)

a hơn b

a hơn b

a hơn b 

chúc học giỏi

Ta có :  \(A=\frac{10^{2016}+1}{10^{2017}+1}\) 

Suy ra  \(10A=\frac{10^{2017}+10}{10^{2017}+1}\) 

Suy ra  \(10A=1+\frac{9}{10^{2017}+1}\) 

Ta lại có : \(B=\frac{10^{2017}+1}{10^{2018}+1}\) 

Suy ra : \(10B=\frac{10^{2018}+10}{10^{2018}+1}\) 

Suy ra : \(10B=1+\frac{9}{10^{2018}+1}\) 

Vì  \(\frac{9}{10^{2017}+1}>\frac{9}{10^{2018}+1}\) 

Nên  \(1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\) 

Suy ra \(10A>10B\) 

Suy ra \(A>B\)

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

vậy A > B

\(A=\frac{10^{2016}+1}{10^{2017}+1}\)

\(A=\frac{10^{2016}+1}{10^{2017}+1}+\frac{10^{2017}+1}{10^{2017}+1}\)

\(A=\frac{10^{2016}+1+10^{2017}+1}{10^{2017}+1}\)

\(A=\frac{10^{2016}+10^{2017}+1+1}{10^{2016}.10+1}\)

\(A=\frac{10^{2016}.\left(1+10\right)+2}{10^{2016}.10+1}\)

\(A=\frac{10^{2016}.11+2}{10^{2016}.10+1}\)

\(A=\frac{11+2}{10+1}\)

\(A=\frac{13}{11}\)(1)

Làm tương tự phần B

Từ 1 và 2 

\(\Leftrightarrow\)\(\frac{13}{11}=\frac{13}{11}\)

\(\Leftrightarrow\)A = B

Nhân cả hai tử của \(A\)và \(B\)với 2 , ta được :

\(10A=10.\left(\frac{10^{2016}+1}{10^{2017}+1}\right)=\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{2^{2017}+1}\)

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

Vì \(1=1;9=9\)

\(\Rightarrow\)Ta so sánh mẫu , ta có:

\(10^{2017}< 10^{2018}\)

\(\Rightarrow10^{2017}+1< 10^{2018}+1\)

\(\Rightarrow1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)

\(\Rightarrow10A>10B\)

Hay \(A>B\)

Anh hiền àaaaaaaaaaaaaaaaaaaaaaaaaa

Tường đây

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 ~