(\(\frac{5}{2014}\)+\(\frac{4}{2015}\)-\(\frac{3}{2016}\)) *(\(\frac{1}{2}\)-\(\frac{1}{3}\)-\(\frac{1}{6}\))
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26 tháng 3 2020
(\(\frac{5}{2014}\)+ \(\frac{4}{2015}\)-\(\frac{3}{2016}\)) . (\(\frac{1}{2}\)-\(\frac{1}{3}\) - \(\frac{1}{6}\))
= ( \(\frac{5}{2014}\)+ \(\frac{4}{2015}\)- \(\frac{3}{2016}\)) . ( \(\frac{3}{6}\)- \(\frac{2}{6}\) - \(\frac{1}{6}\))
= ( \(\frac{5}{2014}\)+ \(\frac{4}{2015}\)- \(\frac{3}{2016}\)) . 0
= 0
3 tháng 10 2017
Xét tử: \(2015+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)
\(=\left(1+1+...+1\right)+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)( trong ngoặc có 2015 số 1 )
\(=\left(1+\frac{2014}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{1}{2015}\right)+1\)
\(=\frac{2016}{2}+\frac{2016}{3}+\frac{2016}{4}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(=2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
Ghép tử và mẫu \(\frac{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}=2016\)
Vậy \(A=2016\)
20 tháng 3 2016
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
27 tháng 7 2017
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vô bài toán được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}\)
22 tháng 12 2016
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)
S
30 tháng 7 2019
\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)
23 tháng 4 2018
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 ~
23 tháng 4 2018
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
(số 2016 tách ra làm 2016 số 1 rồi cộng vào từng phân số, còn dư 1 số viết thành 2017/2017 nghe bạn!!! :)))
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=2017\)
20 tháng 3 2017
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
9 tháng 7 2018
\(B=\frac{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}\)
\(B=\frac{\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}\)
\(B=\frac{\frac{2016}{2}+...+\frac{2016}{2015}+\frac{2016}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}}\)
\(B=\frac{2016\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}\)
\(B=2016\)
AK
9 tháng 7 2018
\(B=\frac{\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+\frac{2012}{4}+...+\frac{1}{2015}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}\)
\(\Rightarrow B=\frac{1+\left(\frac{2014}{2}+1\right)+\left(\frac{2013}{3}+1\right)+\left(\frac{2012}{4}+1\right)+...+\left(\frac{1}{2015}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}\)
\(\Rightarrow B=\frac{\frac{2016}{2016}+\frac{2016}{2}+\frac{2016}{3}+\frac{2016}{4}+...+\frac{2016}{2015}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}\)
\(\Rightarrow B=\frac{2016\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}\)
\(\Rightarrow B=2016\)
Vậy \(B=2016\)