Cho \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\)
\(B=\frac{2015}{1}+\frac{2014}{2}+...+\frac{2}{2014}+\frac{1}{2015}\)
Tính \(\frac{A}{B}\)
Cho :
A = \(\left(\frac{1}{2}_{ }+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.....+\frac{1}{2016}+\frac{1}{2017}\right)\)
B = \(\left(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Tính \(\frac{B}{A}\) ?
[Các bạn giúp mình với !!!]
sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)
\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2014}+\frac{1}{2015}}\)
xét mẫu(chỗ 1/2014 sửa lại thành 2/2014)
=(1/2015+1)+(2/2014+1)+...+(2013/3+1)+(2014/2+1)+(2015/1-2014)
=2016/2015+2016/2014+...+2016/3+2016/2+1
=2016.(1/2016+1/2015+...+1/4+1/3+1/2)
=> A= 1/2016
mún dễ hỉu hơn hãy gửi tin nhắn cho mik
\(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2016}\)\(A=\frac{1}{2015}+\frac{1}{2014}+......+\frac{2014}{2}+2015\)
Tính giá trị A phần B
Cho \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\)
\(B=\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+...+\frac{2}{2014}\)\(+\frac{1}{2015}\)
Tính \(\frac{A}{B}\)
Mong mọi người giúp đỡ
Sao đề lạ dữ vậy bạn kiểm tra lại xem cái phần B ấy
Tính nhanh : \(\frac{2017+\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}}\)
Cho \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}\); \(B=\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}\)
Tính \(\frac{A}{B}\)
Vậy \(\frac{A}{B}=\frac{1}{2017}.\)
Chúc bạn học tốt!
cho A=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{2016}+\frac{1}{2017}\)
va B=\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+......+\frac{2}{2015}+\frac{1}{2016}\)
Tinh ti so \(\frac{A}{B}\)
SO SÁNH:
A=\(\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2016}+\frac{1}{2017}}\)
VÀ
B=2017
Mấy bài dạng này biết cách làm là oke
Ta có :
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\left(2016-1-1-...-1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=2017\)
Vậy \(A=2017\)
Chúc bạn học tốt ~
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
(số 2016 tách ra làm 2016 số 1 rồi cộng vào từng phân số, còn dư 1 số viết thành 2017/2017 nghe bạn!!! :)))
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=2017\)
Cho \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017};B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\).CMR B/A là số nguyên
Ta có :
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
\(B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)
\(B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)
\(B=2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{B}{A}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}=2017\)
Vậy \(\frac{B}{A}\)là số nguyên