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\(\text{đặt}k=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(K=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(K=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1008}\right)\)
\(K=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+....+\frac{1}{2017}\Rightarrow A=1\)
Ta có:
\(\Rightarrow A=B.\)
\(\Rightarrow A^{2017}=B^{2017}\)
\(\Rightarrow\left(A^{2017}-B^{2017}\right)^{2018}=\left(B^{2017}-B^{2017}\right)^{2018}=0^{2018}=0.\)
Vậy \(\left(A^{2017}-B^{2017}\right)^{2018}=0.\)
Chúc bạn học tốt!
Đặt \(A=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.......+\frac{2}{2015}+\frac{1}{2016}\)
\(=\frac{2015}{2}+1+\frac{2014}{3}+1+...........+\frac{1}{2015}+1\)
\(=\frac{2017}{2}+\frac{2017}{3}+.........+\frac{2017}{2015}+\frac{2017}{2016}\)
\(=2017.\left(\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2015}+\frac{1}{2016}\right)\)
Thay A vào biểu thức ta dc
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2017}}{A}\)
\(=\frac{\frac{1}{2017}}{2017}\)\(=1\)
CÓ THỂ LÀ SAI NÊN BẠ THÔNG CẢM CHO MK
\(B=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}=\frac{1+\frac{2015}{2}+1+...+\frac{1}{2016}+1}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)(2016 số hạng 1 ở tử số)
\(=\frac{\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}=2017\)
Đặt A là tên của biểu thức
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}-\left(1+\frac{1}{2}+...+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}+\frac{1}{2017}\)
Do đó \(A=\frac{\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2017}}{\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2017}}=1\)