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Đặt phân thức trên là D
=> D=(1+1+1+1+...+1+2013/2+2012/3+...+2/2013+1/2014)/(1/2+1/3+1/4+...+1/2014)
=> D=(1+2013/2+1+2012/3+1+2011/4+...+1+2/2013+1+1/2014+1)/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=(2015/2+2015/3+2015/4+...+2015/2013+2015/2014+1)/(1/2+1/3+1/4+...+1/2014)
=> D=[2015*(1/2+1/3+1/4+1/5+....+1/2014)]/(1/2+1/3+1/4+1/5+...+1/2014)
=> D=2015
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 \(A=2016\)
$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$
$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$
$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$
$\frac{\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$
$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}}$
Ta có: \(\frac{\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...\frac{1}{2014}+2014}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=
= \(\frac{\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+...+\left(\frac{1}{2014}+1\right)+1+2014}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=
= \(\frac{\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}+2015}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=\(\frac{2015.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+1\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=2015
Xét tử:
\(2012+\frac{2011}{2}+\frac{2010}{3}+\frac{2009}{4}+...+\frac{1}{2012}\)
= \(\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{1}{2012}\right)+1\)
= \(\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}\)
= \(2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)\)
Thay vào ta có:
A = \(\frac{2013\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2013}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}\)
=> A = 2013
Mà 2013 chia hết cho 3
=> A chia hết cho 3
A=\(\frac{1+\frac{2011}{2}+1+\frac{2010}{3}+1+...+\frac{1}{2012}+1+1}{\frac{1}{2}+...+\frac{1}{2013}}\)
A=\(\frac{\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}}{\frac{1}{2}+...+\frac{1}{2013}}\)
A=\(\frac{2013\left(\frac{1}{2}+...+\frac{1}{2013}\right)}{\frac{1}{2}+...+\frac{1}{2013}}\)
A=2013
Mà 2013: 3 = 671
Vậy A : 3 dư 0 hay\(A⋮3\)
mỗi số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15
ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
hay tong tren be hon 15
nhớ phải 4 k thì làm
tớ cần gấp !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!