\(A=\frac{T}{M}\)

\(M=\frac{2012}{2}+1+\frac{2011}{3}+1+.....+\frac{1}{2013}+1=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}\)

     \(=2014\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2013}\right)=2014.T\)

\(A=\frac{T}{M}=\frac{T}{2014.T}=\frac{1}{2014}\)

\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2014}{2}+\frac{2014}{3}+\frac{2014}{4}+...+\frac{2014}{2013}}\)=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)}=\frac{1}{2014}\)

bn xem kết quả có đúng ko?

bấm máy tính ra kết quả ai trả làm được phải làm cách giải mới khó

\(A=1+2+3+4......+2^{2010}\)

\(B=2^{2011-1}\)

\(B=2^{2011-1}=2.2.2.2......2=2^{2010}\)

\(=>A=1+2+3.....+2^{2010}>B=2^{2010}\)

(1+5/4+3/2+............................+19/4):23

=4/4+5/4+6/4+7/4+8/4+.......................+19/4):23

=\(\frac{4+5+6+........+19}{4}\):23

=\(\frac{184}{4}\):23=46:23=2

(1+5/4+3/2+............+19/4):23

=46:23=2

Bùi Anh Tuấn ơi tính làm sao ra 46 vậy bạn ơi

A=1 đoán :))

Mình sẽ giúp  bạn cho

mình giúp cho

Ta có: \(\frac{1}{3^2}< \frac{1}{2.3}\)

           \(\frac{1}{4^2}< \frac{1}{3.4}\)

            .....................

            \(\frac{1}{2014^2}< \frac{1}{2013.2014}\)

\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)

Đặt \(B=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)

           \(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)

             \(=\frac{1}{2}-\frac{1}{2014}< \frac{1}{2}\)

\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2}=\frac{3}{4}\)

\(\text{Ta có: }n^2>n^2-1=\left(n-1\right)\left(n+1\right)\)

\(\Rightarrow\frac{1}{n^2}< \frac{1}{\left(n-1\right)\left(n+1\right)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2014^2}< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+...+\frac{1}{2}\left(\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{2014}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(\frac{3}{2}-\frac{1}{2014}-\frac{1}{2015}\right)\)

\(=\frac{3}{4}-\frac{1}{2}\left(\frac{1}{2014}+\frac{1}{2015}\right)< \frac{3}{4}\)

Vậy .............

\(x+\frac{9}{4}=\frac{1}{3}+\frac{1}{4}-\frac{1}{-5}-\left(-\frac{2}{3}\right)-\left(-\frac{1}{4}\right)+\frac{4}{5}\)

\(x+\frac{9}{4}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{3}+\frac{1}{4}+\frac{4}{5}\)

\(x+\frac{9}{4}=\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{4}{5}\right)\)

\(x+\frac{9}{4}=1+\frac{1}{2}+1\)

\(x+\frac{9}{4}=\frac{5}{2}\)

\(x=\frac{5}{2}-\frac{9}{4}\)

\(x=\frac{1}{4}\)

TL:

1/2+1/3=3/6+2/6=5/6

HT

1/2+ 1/3 =5/6 nhé.

\(\frac{1}{2}\)\(+\)\(\frac{1}{3}\)\(=\)\(\frac{3}{6}\)\(+\)\(\frac{2}{6}\)\(=\)\(\frac{5}{6}\)