Xét N ta có :

N = \(\frac{-7}{10^{2005}}\)+ \(\frac{-15}{10^{2006}}\)

N = \(\frac{-7}{10^{2005}}\)+ \(\frac{-7}{10^{2006}}\)+\(\frac{-8}{10^{2006}}\)

Xét M ta có :

M = \(\frac{-15}{10^{2005}}\)+\(\frac{-7}{10^{2006}}\)

M = \(\frac{-8}{10^{2005}}\)+\(\frac{-7}{10^{2005}}\)+ \(\frac{-7}{10^{2006}}\)

Vì \(\frac{-8}{10^{2006}}\)< \(\frac{-8}{10^{2005}}\) => N < M

Khó V~

ta có:M-N=\(\dfrac{-15}{10^{2005}}-\dfrac{-7}{10^{2005}}+\dfrac{-7}{10^{2006}}-\dfrac{-15}{10^{2006}}\)

=\(\dfrac{-8}{10^{2015}}\)+\(\dfrac{8}{10^{2016}}\)=8(\(\dfrac{1}{10^{2006}}-\dfrac{1}{10^{2005}}\))

Ta có:10²⁰⁰⁶>10²⁰⁰⁵<=>\(\dfrac{1}{10^{2006}}< \dfrac{1}{10^{2005}}\)

<=>\(\dfrac{1}{10^{2006}}-\dfrac{1}{10^{2005}}< 0\)

=>M-N<0 hay N>M

Giải 

\(\frac{-7}{10^{2005}}+\frac{-15}{10^{2016}}=\frac{-7.10}{10^{2005}.10}+\frac{-15}{10^{2006}}=\frac{-70}{10^{2006}}+\frac{-15}{10^{2006}}=\frac{-85}{10^{2006}}\left(1\right).\)

\(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}=\frac{-15.10}{10^{2005}.10}+\frac{-7}{10^{2006}}=\frac{-150}{10^{2006}}+\frac{-7}{10^{2006}}=-\frac{157}{10^{2006}}\left(2\right).\)

Từ (1) và ( 2 ) => (1) > (2)

học tốt

Xét A ta có

         A=\(\frac{-7}{10^{2005}}\) + \(\frac{-15}{10^{2006}}\)

        A=\(\frac{-7}{10^{2005}}\) +\(\frac{-8}{10^{2006}}\) +\(\frac{-7}{10^{2006}}\)

Xét B ta có

     B=\(\frac{-15}{10^{2005}}\) +\(\frac{-7}{10^{2006}}\)

     B=\(\frac{-8}{10^{2005}}\) + \(\frac{-7}{10^{2005}}\) +\(\frac{-7}{10^{2006}}\)

Vì \(\frac{-8}{10^{2006}}\) >\(\frac{-8}{10^{2005}}\) nên A>B

A>B mk chắc chắn

Làm như bạn Nguyễn Vũ Phượng Thảo, nhưng dấu giữa A và B là ngược lại. A<B

\(N=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\)

\(M=\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)

Ta xét M và N, ta có: \(\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}\text{ chung}\)

Mà: \(\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\Rightarrow M>N\)

Ta có

\(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\)

\(B=\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)

Vì \(\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\)

=>A>B

\(A=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\)

\(B=\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)

Vì \(\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\) 

\(\Rightarrow A>B\)

A > B.

Tích nha bạn !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

hình như có nhầm chút thì phải?

-8/10^2006 < -8/10^2005 chứ!!!