Chào bạn, bạn hãy theo dõi câu trả lời của mình nhé! 

a) Ta có : 

\(2^{225}=\left(2^3\right)^{75}=8^{75}\)

\(3^{151}=3^{150}\cdot3=\left(3^2\right)^{75}\cdot3=9^{75}\cdot3\)

Mà \(9^{75}>8^{75}=>9^{75}\cdot3>8^{75}=>3^{151}>2^{225}\)

b) Nhân cả vế A lẫn vế B với 10²⁰⁰⁵, ta có : 

\(10^{2005}A=-7+\frac{-15}{10}=\frac{-70}{10}+\frac{-15}{10}=\frac{-85}{10}\)

\(10^{2005}B=-15+\frac{-7}{10}=\frac{-150}{10}+\frac{-7}{10}=\frac{-157}{10}\)

Mà \(\frac{-85}{10}>\frac{-157}{10}=>10^{2005}A>10^{2005}B\)

\(=>A>B\)

Chúc bạn học tốt!

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~

A=\(\frac{2005^{2005}+1}{2005^{2006}+1}\) < 1 => \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < \(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\) = \(\frac{2005^{2005}+2005}{2005^{2006}+2005}\)= \(\frac{2005.\left(2005^{2004}+1\right)}{2005.\left(2005^{2005}+1\right)}\) = \(\frac{2005^{2004}+1}{2005^{2005}+1}\) = B => A<B.

Ta thấy:A=\(\frac{2005^{2005+1}}{2005^{2006}+1}\)<1
Ta có:A=\(\frac{2005^{2005}+1}{2005^{2006}+1}\)<\(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\)=\(\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\)=b
Vậy A<B
Chắc chắn 100%

Ta có :

Tử số = \(\frac{2006}{2}+...+\frac{2006}{2007}\)

= 2006.(\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}\))

MS= \(\frac{2006}{1}+\frac{2005}{2}+...+\frac{1}{2006}\)

= 2006+\(\frac{2007-2}{2}+\frac{2007-3}{3}+...+\frac{2007-2006}{2006}\)

=200+.(\(\frac{2007}{2}+\frac{2007}{3}+...+\frac{2007}{2006}\)) - ( 1+1+1+...+1 )(2006c/s1) 

= 2006 . (\(\frac{2007}{2}+...+\frac{2007}{2006}\))-2006

=\(\frac{2007}{2}+...+\frac{2007}{2006}\)

=2007.(\(\frac{1}{2}+...+\frac{1}{2006}\))

Khi đó : 

C= .... bạn tự đáp số 

và cuối cùng C = \(\frac{2006}{2007}\)

a) Ta có:  \(\frac{-9}{80}=\frac{\left(-9\right)x4}{80x4}=\frac{-36}{320}\) và \(\frac{17}{320}\)

b) Ta có:  \(\frac{-7}{10}=\frac{\left(-7\right)x33}{10x33}=\frac{-231}{330}\) và \(\frac{1}{33}=\frac{1x10}{33x10}=\frac{10}{330}\)

c) Ta có:

\(\frac{-5}{14}=\frac{\left(-5\right)x10}{14x10}=\frac{-50}{140}\)

\(\frac{3}{20}=\frac{3x7}{20x7}=\frac{21}{140}\)

\(\frac{9}{70}=\frac{9x2}{70x2}=\frac{18}{140}\)

d) Ta có: 

\(\frac{10}{42}=\frac{10x22}{42x22}=\frac{220}{924}\)

\(\frac{-3}{28}=\frac{\left(-3\right)x33}{28x33}=\frac{-99}{924}\)

\(\frac{-55}{132}=\frac{\left(-55\right)x7}{132x7}=\frac{-385}{924}\)

sai đề

\(A=\frac{10^8+2}{10^8-1}=\frac{\left(10^8-1\right)+3}{10^8-1}=\frac{10^8-1}{10^8-1}+\frac{3}{10^8-1}=1+\frac{3}{10^8-1}\)

\(B=\frac{10^8}{10^8-3}=\frac{\left(10^8-3\right)+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)

Vì \(1+\frac{3}{10^8-1}<1+\frac{3}{10^8-3}\) nên A < B

Ta có : 

 A = 10⁸ + 2 / 10 ⁸ - 1 = 1 + 3 / 10 ⁸ - 1

B = 10⁸ / 10 ⁸ - 3 =  1 + 3 / 10⁸ - 3

Vì 3/ 10⁸ - 1 < 3 / 10⁸ - 3=> A < B

B=\(\frac{2011^{10}-1}{2011^{10}-3}\) <1 => \(\frac{2011^{10}-1}{2011^{10}-3}\) < \(\frac{2011^{10}-1+2}{2011^{10}-3+2}\) = \(\frac{2011^{10}+1}{2011^{10}-1}\) = A

=> B<A

Cảm ơn bạn nhiều nha giải ra lại thấy dễ ak