Ta có: `8^111 =(2^3 )^111 =2^(3.111)=2^333`

`4^170 =(2^2 )^170 =2^(2.170)=2^340`

Vì `333<340=>8^111 <4^170`

Ta có: `3^300 =3^(3.100)=(3^3 )^100=27^100`

`5^200 =5^(2.100)=(5^2 )^100 =25^100`

Vì `27>25=>3^300 >5^200`

a: 8^111=2^333

4^170=(2^2)^170=2^340

mà 333<340

nên 8^111<4^170

b: 3^300=(3^3)^100=27^100

(5^200)=(5^2)^100=25^100

mà 27>25

nên 3^300>5^200

DẶT A= BIỂU THỨC TRÊN

A=2+1+1+..+1-(1/4+1/9+...+1/2500)

ĐẶT S=1/4+1/9+...+1/2500

S=1/2^2+1/3^2+...+1/50^2

SÓ SỐ HẠNG CỦA S:

(50-2)/1+1=49

SUY RA 

1+1+...+1=49

SUY RA A=2+49-S

A=51-S

TAO CÓ :

S<1/1.2+1/2.3+...+1/49.100

S<1-1/2+1/2-1/3+...+1/49-1/50

S<1-1/50

S<49/50

SUY RA A>51-49/50

SUY RA A>50

\(A=2+\frac{3}{4}+\frac{8}{9}+......+\frac{2499}{2500}\)

\(A=2+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+.....+\left(1-\frac{1}{2500}\right)\)

\(A=2+1-\frac{1}{4}+1-\frac{1}{9}+.........+1-\frac{1}{2500}\)

\(A=2+\left(1+1+....+1\right)-\left(\frac{1}{4}+\frac{1}{9}+....+\frac{1}{2500}\right)\)

\(A=2+\left(1+1+....+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{50^2}\right)\)

Vì mỗi số 1 đều đi với 1 phân số nên có số số 1 là: (50-1)/1+1=50(số)

\(A=52-\left(\frac{1}{2^2}+\frac{1}{3^2}+.......+\frac{1}{50^2}\right)\)

\(\frac{1}{2^2}<\frac{1}{1\cdot2}\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}\)

.........

\(\frac{1}{50^2}<\frac{1}{49\cdot50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{49\cdot50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{1}{1}-\frac{1}{50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{49}{50}\)

\(\Rightarrow52-\left(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}\right)>52-\frac{49}{50}\)

\(\Rightarrow A>51\frac{1}{50}\)

Vì\(51\frac{1}{50}>50\Rightarrow A>50\)