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Đặt : \(A=\frac{2018^{13}+1}{2018^{14}+1}\); \(B=\frac{2018^{2012}+1}{2018^{2013}+1}\)
Ta có :
\(2018A=\frac{2018.\left(2018^{13}+1\right)}{2018^{14}+1}\)
\(2018A=\frac{2018^{14}+2018}{2018^{14}+1}=\frac{2018^{14}+1+2017}{2018^{14}+1}=\frac{2018^{2014}+1}{2018^{14}+1}+\frac{2017}{2018^{14}+1}=1+\frac{2017}{2018^{14}+1}\)
\(2018B=\frac{2018.\left(2018^{12}+1\right)}{2018^{13}+1}\)
\(2018B=\frac{2018^{13}+2018}{2018^{13}+1}=\frac{2018^{13}+1+2017}{2018^{13}+1}=\frac{2018^{13}+1}{2018^{13}+1}+\frac{2017}{2018^{13}+1}=1+\frac{2017}{2018^{13}+1}\)
Vì 201814 + 1 > 201813 + 1 nên \(\frac{2017}{2018^{14}+1}< \frac{2017}{2018^{13}+1}\)
\(\Rightarrow1+\frac{2017}{2018^{14}+1}< 1+\frac{2017}{2018^{13}+1}\)Hay : A < B
Vậy A < B
Đặt \(A=\frac{2018^{13}+1}{2018^{14}+1}\)và \(B=\frac{2018^{12}+1}{2018^{13}+1}\)
Ta có :
\(2018A=\frac{\left(2018^{13}+1\right)\times2018}{2018^{14}+1}\) \(2018B=\frac{\left(2018^{12}+1\right)\times2018}{2018^{13}+1}\)
\(2018A=\frac{2018^{14}+2018}{2018^{14}+1}\) \(2018B=\frac{2018^{13}+2018}{2018^{13}+1}\)
\(2018A=\frac{2018^{14}+1+2017}{2018^{14}+1}\) \(2018B=\frac{2018^{13}+1+2017}{2018^{13}+1}\)
\(2018A=1+\frac{2017}{2018^{14}+1}\) \(2018B=1+\frac{2017}{2018^{13}+1}\)
Vì \(\frac{2017}{2018^{14}+1}< \frac{2017}{2018^{13}+1}\)
\(\Rightarrow2018A< 2018B\)
\(\Rightarrow A< B\)
Vậy : \(\frac{2018^{13}+1}{2018^{14}+1}< \frac{2018^{12}+1}{2018^{13}+1}\)
\(A=\frac{10^{2016}+2018}{10^{2017}+2018}\)
\(\Rightarrow10A=\frac{10^{2017}+20180}{10^{2017}+2018}\)
\(=\frac{10^{2017}+2018+18162}{10^{2017}+2018}\)
\(=\frac{10^{2017}+2018}{10^{2017}+2018}+\frac{18162}{10^{2017}+2018}\)
\(=1+\frac{18162}{10^{2017}+2018}\)
\(B=\frac{10^{2017}+2018}{10^{2018}+2018}\)
\(\Rightarrow10B=\frac{10^{2018}+20180}{10^{2018}+2018}\)
\(=\frac{10^{2018}+2018+18162}{10^{2018}+2018}\)
\(=\frac{10^{2018}+2018}{10^{2018}+2018}+\frac{18162}{10^{2018}+2018}\)
\(=1+\frac{18162}{10^{2018}+2018}\)
Ta thấy: \(1+\frac{18162}{10^{2017}+2018}>1+\frac{18162}{10^{2018}+2018}\)
=> 10A > 10B
=> A > B
\(M=\frac{10^{2018}+2}{10^{2018}+1}=\frac{10^{2018}+1+1}{10^{2018}+1}=\frac{10^{2018}+1}{10^{2018}+1}+\frac{1}{10^{2018}+1}=1+\frac{1}{10^{2018}+1}\)
\(N=\frac{10^{2018}}{10^{2018}-3}=\frac{10^{2018}-3+3}{10^{2018}-3}=\frac{10^{2018}-3}{10^{2018}-3}+\frac{3}{10^{2018}-3}=1+\frac{3}{10^{2018}-3}\)
Ta có: \(\frac{1}{10^{2018}+1}< \frac{1}{10^{2018}-3}< \frac{3}{10^{2018}-3}\)
\(\Rightarrow N>M\)
\(M=\frac{10^{2018}+2}{10^{2018}+1}=\frac{10^{2018}+1+1}{10^{2018}+1}=\frac{10^{2018}+1}{10^{2018}+1}+\frac{1}{10^{2018}+1}=1+\frac{1}{10^{2018}+1}.\)
\(N=\frac{10^{2018}}{10^{2018}-3}=\frac{10^{2018}-3+3}{10^{2018}-3}=\frac{10^{2018}-3}{10^{2018}-3}+\frac{3}{10^{2018}-3}=1+\frac{3}{10^{2018}-3}\)
Ta có\(\frac{1}{10^{2018}+1}< \frac{1}{10^{2018}-3}< \frac{3}{10^{2018}-3}\)
\(\Leftrightarrow N>M\)
\(+)A=\frac{10^{2016}+2018}{10^{2017}+2018}\)
\(10A=\frac{10^{2017}+20180}{10^{2017}+2018}=1+\frac{18162}{10^{2017}+2018}\left(1\right)\)
\(+)10B=\frac{10^{2018}+20180}{10^{2018}+2018}=1+\frac{18162}{10^{2018}+2018}\left(2\right)\)
Từ (1),(2)=> \(\frac{18162}{10^{2017}+2018} >\frac{18162}{10^{2018}+2018}\)
=> 10A>10B
=>A>B
\(M=\frac{2018^{2018}+1}{2019^{2019}+1}\)
\(\Leftrightarrow2M=1+\frac{2017}{2018^{2019}+1}\)
\(N=\frac{2018^{2019}-2}{2018^{2020}-2}\)
\(\Leftrightarrow2N=1-\frac{4034}{2018^{2020}-2}\)
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:)))
\(2018^{13}-2018^{12}=2018^{12}\left(2018-1\right)=2018^{12}.2017\)
\(2018^{11}.2018^{10}=2018^{12}.2018^9\)
Nhận thấy: \(2017< 2018^9\)=> \(2018^{12}.2017< 2018^{12}.2018^9\)
hay \(2018^{13}-2018^{12}< 2018^{11}.2018^{10}\)
Mik đang nghĩ là vậy chứ chắc giải thik ko đúng đâu...
\(2018^{13}-2018^{12}< 2018^{11}2018^{10}\)
Vì : Phép tính \(2018^{13}-2018^{12}\) đã trừ đi thì chỉ còn một số nhỏ hơn phép tính \(2018^{11}2018^{10}\)
Mik nghĩ thôi nhé, chắc ko đúng đâu
k cho mik nhé bn