A = 6cs + 7cs - 1 = 7cs              

B = 12cs - 2 = 12 cs

==>A>B

\(M=\frac{2018^{2018}+1}{2019^{2019}+1}\)

\(\Leftrightarrow2M=1+\frac{2017}{2018^{2019}+1}\)

\(N=\frac{2018^{2019}-2}{2018^{2020}-2}\)

\(\Leftrightarrow2N=1-\frac{4034}{2018^{2020}-2}\)

Nhận thấy :  \(1+\frac{2017}{2018^{2019}+1}>1-\frac{4034}{2018^{2020}-2}\Leftrightarrow2M>2N\Leftrightarrow M>N\)

Từ đề bài, ta suy ra:

So sánh hai biểu thức

\(M=\left(2018^{2018}+1\right)\cdot\left(2018^{2020}-2\right)\)(1)

\(N=\left(2018^{2019}-2\right)\cdot\left(2018^{2019}+1\right)\)(2)

Xét biểu thức M và N, ta suy ra:

\(M=\left(2018^{2019}-2017\right)\cdot\left(2019^{2019}+2016\right)\)

\(N=\left(2018^{2019}-2017\right)\cdot\left(2018^{2018}-2016\right)\)

Nhận thấy (2019²⁰¹⁹+2016)>(2018²⁰¹⁸-2016) nên M>N

Vậy M>N.

P/s:Mình đây không phải top 10 tuần nên bài có thể sai sót, mong bạn tham khảo:)))

\(M=\frac{10^{2018}+1}{10^{2019}+1}\)

\(\Rightarrow10M=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)

\(N=\frac{10^{2019}+1}{10^{2020}+1}\)

\(\Rightarrow10N=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)

Ta co: \(\frac{9}{10^{2019}+1}>\frac{9}{10^{2020}+1}\) ma \(1=1\)

\(\Rightarrow1+\frac{9}{10^{2019}+1}>1+\frac{9}{10^{2020}+1}\)

\(\Rightarrow10M>10N\)

\(\Rightarrow M>N\)

\(+)A=\frac{10^{2016}+2018}{10^{2017}+2018}\)

\(10A=\frac{10^{2017}+20180}{10^{2017}+2018}=1+\frac{18162}{10^{2017}+2018}\left(1\right)\)

\(+)10B=\frac{10^{2018}+20180}{10^{2018}+2018}=1+\frac{18162}{10^{2018}+2018}\left(2\right)\)

Từ (1),(2)=> \(\frac{18162}{10^{2017}+2018} >\frac{18162}{10^{2018}+2018}\)

=> 10A>10B

=>A>B

k đúng cho mình đi, mình giải cho.

\(A=\frac{10^{2016}+2018}{10^{2017}+2018}\)

\(\Rightarrow10A=\frac{10^{2017}+20180}{10^{2017}+2018}\)

\(=\frac{10^{2017}+2018+18162}{10^{2017}+2018}\)

\(=\frac{10^{2017}+2018}{10^{2017}+2018}+\frac{18162}{10^{2017}+2018}\)

\(=1+\frac{18162}{10^{2017}+2018}\)

\(B=\frac{10^{2017}+2018}{10^{2018}+2018}\)

\(\Rightarrow10B=\frac{10^{2018}+20180}{10^{2018}+2018}\)

\(=\frac{10^{2018}+2018+18162}{10^{2018}+2018}\)

\(=\frac{10^{2018}+2018}{10^{2018}+2018}+\frac{18162}{10^{2018}+2018}\)

\(=1+\frac{18162}{10^{2018}+2018}\)

Ta thấy: \(1+\frac{18162}{10^{2017}+2018}>1+\frac{18162}{10^{2018}+2018}\)

=> 10A > 10B

=> A > B

ta có :

\(A=\frac{10^{2019}+1}{10^{2018}+1}=\frac{10^{2018}.10+1}{10^{2018}+1}=\frac{10}{10^{2018}+1}\)

\(B=\frac{10^{2018}+1}{10^{2017}+1}=\frac{10^{2017}.10+1}{10^{2017}+1}=\frac{10}{10^{2017}+1}\)

Do \(10^{2017}+1< 10^{2018}+1\Rightarrow\frac{10}{10^{2017}+1}>\frac{10}{10^{2018}+1}\)

\(\Rightarrow A< B\)

ta có quy đồng B ta dc(-9x10^2018-19x10^2019)/(10^2019x10^2018)

tương tự với C ta có (-19x10^2018-9x10^2019)/(10^2019x10^2018)

sau khi quy đồng ta thấy mẫu của B và C giống nhau từ đó ta so sánh tử số của B và C

tử số của B=10^2018x(-9-19x10)=10^2018x-199

C=10^2018x(-19-9x10)=10^2018x-109

ta thấy -199<-109=>B<C (dpcm)