Ta có:

\(\frac{1}{2}< \frac{2}{3}\)

\(\frac{3}{4}< \frac{4}{5}\)

\(\frac{5}{6}< \frac{6}{7}\)

\(...\)

\(\frac{99}{100}< \frac{100}{101}\)

\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)

\(\Rightarrow M< N\)

+1 với +5 bên A bằng với bên B nên ta chỉ so sánh 126^99/12^100 bên A và 12^100/12^101 bên B

Vì 12^99[A]<12^100[B] và 12^100[A]<12^101[B] 

=>A<B

2:

a: A=1+2+2^2+2^3+2^4

=>2A=2+2^2+2^3+2^4+2^5

=>A=2^5-1

=>A=B

b: C=3+3^2+...+3^100

=>3C=3^2+3^3+...+3^101

=>2C=3^101-3

=>\(C=\dfrac{3^{101}-3}{2}\)

=>C=D

Ta có: 

\(\left\{\begin{matrix}5^{27}=\left(5^3\right)^9=125^9\\2^{63}=\left(2^7\right)^9=128^9\end{matrix}\right\}\Rightarrow5^{27}< 2^{63}\left(1\right)\)

\(\left\{\begin{matrix}2^{63}=\left(2^9\right)^7=512^7\\5^{28}=\left(5^4\right)^7=625^7\end{matrix}\right\}\Rightarrow2^{63}< 5^{28}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow5^{27}< 2^{63}< 5^{28}\) (đpcm)

ai bt tự làm

ngu tự chịu

       A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101

=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4

=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)

=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101

=> 4A = 99*100*101*102

=> 4A = 101989800

=>   A = 25497450

a)A=3^0+3^1+3^2+3^3+...+3^2012

A=1+3+3^2+3^3+..+3^2012

3A=3+3^2+3^3+3^4+..+3^2013

3A-A=3+3^2+3^3+3^4+..+3^2013-1-3-3^2-3^3-...-3^2012

2A=3^2013-1

A=\(\frac{3^{2013}-1}{2}\)

B=3^2013

=> A>B

b) A=1+5+5^2+5^3+..+5^99+5^100

5A=5+5^2+5^3+5^4+...+5^100+5^101

5A-A=5+5^2+5^3+5^4+..+5^100+5^101-1-5-5^2-5^3-..-5^99-5^100

4A=5^101-1

A=\(\frac{5^{101}-1}{4}\)

B=5^101/4

=> A<B

nhân 3A lên

nhân 5B lên