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\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)
=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)
\(A=\left(1-\frac{1}{2014}\right)\left(1-\frac{2}{2014}\right)......\left(1-\frac{2015}{2014}\right)\)
\(=\left(1-\frac{1}{2014}\right)\left(1-\frac{2}{2014}\right).....\left(1-\frac{2014}{2014}\right)\left(1-\frac{2015}{2014}\right)\)
\(=\left(1-\frac{1}{2014}\right)\left(1-\frac{2}{2014}\right)......0.\left(1-\frac{2015}{2014}\right)\)
\(=0\)
Lời giải:
$A=2015+\frac{2015}{1+2}+\frac{2015}{1+2+3}+...+\frac{2015}{1+2+3+...+2014}$
$=2015+\frac{2015}{\frac{2.3}{2}}+\frac{2015}{\frac{3.4}{2}}+....+\frac{2015}{\frac{2014.2015}{2}}$
$=2015+4030(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015})$
$=2015+4030(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015})$
$=2015+4030(\frac{1}{2}-\frac{1}{2015})=2015+2015-2$
$=4028$
\(B=\left(\dfrac{1}{2015}+1\right)+\left(\dfrac{2}{2014}+1\right)+\left(\dfrac{3}{2013}+1\right)+...+\left(\dfrac{2014}{2}+1\right)+1\)
\(=\dfrac{2016}{2}+\dfrac{2016}{3}+...+\dfrac{2016}{2016}\)
=>B:A=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
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2015}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2014}{2015}\)
\(=\frac{1}{2015}< \frac{1}{2014}\).