Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}\)\
\(A=\frac{1}{2014}\)
B=\(\left[\left(\frac{1}{3}+\frac{1}{4}\right)x\frac{12}{19}+\frac{12}{19}\right]:\frac{4}{5}-\frac{1}{4}+2012\)
B=\(\left(\frac{7}{12}x\frac{12}{19}+\frac{12}{19}\right):\frac{4}{5}-\frac{1}{4}+2012\)
B=\(\left(\frac{7}{19}+\frac{12}{19}\right):\frac{4}{5}-\frac{1}{4}+2012\)
B=\(\frac{5}{4}-\frac{1}{4}+2012\)
B=1+2012
B=2013
\(B=[\left(\frac{1}{3}+\frac{1}{4}\right)\times\frac{12}{19}+\frac{12}{19}]:\frac{4}{5}-\frac{1}{4}+2012\)
\(B=[\frac{7}{12}\times\frac{12}{19}+\frac{12}{19}]:\frac{4}{5}-\frac{1}{4}+2012\)
\(B=[\frac{7}{19}+\frac{12}{19}]:\frac{4}{5}-\frac{1}{4}+2012\)
\(B=1:\frac{4}{5}-\frac{1}{4}+2012\)
\(B=\frac{5}{4}-\frac{1}{4}+2012\)
\(B=1+2012\)
\(B=2013\)
a=100+98+96+...+2-97-95-...-1
=100+(98-97)+(96-95)+...+(2-1)
=100+1+1+...+1
=100+1.50
=100+50=150
A=1-2+3-4+5-6+....+99-100
A=(1-2)+(3-4)+(5-6)+...+(99-100)
A=-1+(-1)+(-1)+...+(-1) (50 số hạng)
A=-1*50
A=-50
dễ thôi bài giải như này nha 1-2+3-4....+99-100=<-1><-1>....<-1>=-1.50=-50
\(\frac{\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2014}}{-\frac{2013}{1}-\frac{2012}{2}-\frac{2011}{3}-...-\frac{1}{2013}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}{-\left(2013+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2014}}{-\left(\frac{2014}{2013}+\frac{2014}{2}+\frac{2014}{3}+....+\frac{2014}{2013}\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2014}}{-2014\left(\frac{1}{2014}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2013}\right)}\)
\(=-\frac{1}{2014}\)