Ta có :

\(2^{100}=2^{10.10}=\left(2^{10}\right)^{10}=1024^{10}\)

\(1024^{10}>1000^{10}\) \(\left(1\right)\)

Mặt khác :

\(10^{30}=\left(10^3\right)^{10}=1000^{10}\) \(\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra \(10^{30}< 2^{100}\)

Hay \(A< B\)

Mình nói lí thuyết cho nghe:

 Với phân số \(\frac{a-b}{a.b}\)\(\left(VD:\frac{1}{1.2};\frac{1}{2.3};\frac{1}{2015.2016};\frac{3}{15.18};\frac{3}{18.21};\frac{1}{10.11};\frac{1}{11.12};...\right)\)thì:

 \(\frac{b-a}{a.b}=\frac{b}{a.b}-\frac{a}{a.b}=\frac{1}{a}-\frac{1}{b}\left(VD:\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2};\frac{3}{15.18}=\frac{1}{15}-\frac{1}{18}\right)\)

ÁP dụng để tính:

 c) \(\Rightarrow\frac{1}{4}C=\frac{1}{4}\left(\frac{12}{15.18}+\frac{12}{18.21}+...+\frac{12}{87.90}\right)=\frac{3}{15.18}+\frac{3}{18.21}+....+\frac{3}{87.90}\)

\(\Rightarrow\frac{1}{4}C=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}=\frac{1}{15}-\frac{1}{90}\)

=> \(C=\left(\frac{1}{15}-\frac{1}{90}\right).4\)