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Xét: \(\frac{\left(17^{2017}+16^{2017}\right)^{2018}}{17^{2017.2018}}=\left(\frac{17^{2017}+16^{2017}}{17^{2017}}\right)^{2018}=\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}\)
\(\frac{\left(17^{2018}+16^{2018}\right)^{2017}}{17^{2017.2018}}=\left(\frac{17^{2018}+16^{2018}}{17^{2018}}\right)^{2017}=\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)
Ta có: \(0< \frac{16}{17}< 1\)
=> \(\left(\frac{16}{17}\right)^{2017}>\left(\frac{16}{17}\right)^{2018}\)
=> \(1+\left(\frac{16}{17}\right)^{2017}>1+\left(\frac{16}{17}\right)^{2018}>1\)
=> \(\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}>\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)
=> \(\left(17^{2017}+16^{2017}\right)^{2018}>\left(17^{2018}+16^{2018}\right)^{2017}\)
Nhận xét: \(|x-2017|^{2017}\ge0;\left(2y+2018\right)^{2018}=\left(\left(2y+2018\right)^{1009}\right)^2\ge0\)
Tổng của 2 số dương bằng 0 khi và chỉ khi cả 2 số đều bằng 0
=> \(\hept{\begin{cases}|x-2017|^{2017}=0\\\left(2y+2018\right)^{2018}=0\end{cases}}\) <=> \(\hept{\begin{cases}x-2017=0\\2y+2018=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=2017\\y=-1009\end{cases}}\)
Đáp số: (x,y)=(2017; -1009)
Đánh giá: \(\left|x-2017\right|^{2017}\ge0\)
\(\left(2y+2018\right)^{2018}\ge0\)
\(\Rightarrow\)\(\left|x-2017\right|^{2017}+\left(2y+2018\right)^{2018}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x-2017=0\\2y+2018=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=2017\\y=-1009\end{cases}}\)
Vậy,...
Đặt \(A=\frac{2^{2017}+1}{2^{2018}+1}\Rightarrow2A=\frac{2^{2018}+2}{2^{2018}+1}=\frac{2^{2018}+1+1}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)
\(B=\frac{2^{2018}+1}{2^{2019}+1}\Rightarrow2B=\frac{2^{2019}+2}{2^{2019}+1}=\frac{2^{2019}+1+1}{2^{2019}+1}=1+\frac{1}{2^{2019}+1}\)
Vì \(2^{2019}+1>2^{2018}+1\Rightarrow\frac{1}{2^{2019}+1}< \frac{1}{2^{2018}+1}\)
\(\Rightarrow2A>2B\Rightarrow A>B\)
\(M=\left(2018^{2019}+2018^{2018}+...+2018^2+2018\right)2017+1\)
Gọi \(A=2018^{2019}+2018^{2018}+...+2018^2+2018\)
\(\Rightarrow2018A=2018^{2020}+2018^{2019}+...+2018^3+2018^2\)
\(\Rightarrow2018A-A=2018^{2020}-2018\)
\(\Rightarrow2017A=2018^{2020}-2018\)
\(\Rightarrow A=\left(2018^{2020}-2018\right)\div2017\)
\(\Rightarrow M=\left(2018^{2020}-2018\right)\div2017.2017+1\)
\(\Rightarrow M=2018^{2020}-2018+1\)
\(\Rightarrow M=2018^{2020}-2017\)
So Sánh: A=B
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