\(4A=4+4^2+...+4^{100}\)

\(A=1+4+4^2+..+4^{99}\)

\(\Rightarrow3A=4A-A=4^{100}-1\)

\(\Rightarrow3A< 4^{100}\)

\(\Rightarrow\frac{3A}{B}< 1\Rightarrow\frac{A}{B}< \frac{1}{3}\)

use máy tính casio fx 570VN ta đc   x= -100

mình viết nhầm=)))))

\(b,\frac{10}{99}\)+\(\frac{11}{199}\)+\(\frac{12}{299}\).\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{-1}{6}\)

\(A=1-2+3-4+...+97-98+99\)
\(=\left(1-2\right)+\left(3-4\right)+...+\left(97-98\right)+99\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+103\)
  Có \(\left[\left(98-1\right):1+1\right]:2=49\)(cặp số \(\left(-1\right)\))
\(=49.\left(-1\right)+99=-49+99=50\)
\(B=\left(1-4\right)+\left(7-10\right)+...+\left(97-100\right)+103\)
\(=\left(-3\right)+\left(-3\right)+...+\left(-3\right)+103\)
Có \(\left[\left(100-1\right):3+1\right]:2=17\)(cặp số \(\left(-1\right)\))
\(=17.\left(-3\right)+103=\left(-51\right)+103=52\)
=) \(\frac{A}{B}=\frac{50}{52}=\frac{25}{26}\)

49/34

\(A=1+4+4^2+...+4^{99}\)(1)

=>\(4A=4+4^2+4^3+...+4^{100}\)(2)

Lấy (2)-(1) ta được 

3A=4¹⁰⁰-1

=>A=\(\frac{4^{100}-1}{3}<\frac{4^{100}}{3}=B\)

=>A<B (đpcm)

lộn 4^100/3=B/3

=>A<B/3(đpcm)