a: A=16*16+16-5=16*16+11

=>A=B

b: A=2000*2000=2000^2

B=1998*2002=(2000-2)(2000+2)=2000^2-4

=>A<B

d: B=(2020-1)(2020+1)=2020^2-1

=>B<A

(a + b) + c = [2 + (-4)] + (-6) = [ -(4 - 2)] + (-6) = (-2) + (-6) = - (2 + 6) = -8

a + (b + c) = 2 + [(-4) + (-6)] = 2 + [ -(4 + 6)] = 2 + (-10)  = - (10 - 2)= -8

Vậy (a + b) + c = a + (b + c)

a) A < B

b) A < B

a) A<B

B) A<B

\(A=2021^3=2021\cdot2021^2>2021\left(2021^2-1\right)=2021\left(2021\cdot2021-2021+2021-1\right)=2021\cdot\left(2021+1\right)\left(2021-1\right)=2021\cdot2022\cdot2020=B\)

\(B=2020.2021.2022=\left(2021-1\right).2021.\left(2021+1\right)=\left[\left(2021-1\right)\left(\left(2021+1\right)\right)\right].2011=\left(2021^2-1\right).2021=2021^3-2021< 2021^3=A\)

A=1987657x1987655=(1987656+1)x(1987656-1)

=1987656²+1987656-1987656-1

=1987656x1987656-1<1987656x1987656.

Vậy A<B

Ta có:

A=1987657.1987655

A=(1987656+1).1987655

A=1987656.1987655+1987655

B=1987656.1987656

B=(1987655+1).1987656

B=1987655.1987656+1987656

Ta thấy 1987656.1987655+1987655<1987655.1987656+1987656

=>  A<B(1 đơn vị)

a)Ta có:

 A= 1/1.2+1/2.3+1/3.4+.....+1/99.100

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100

=99/100

b)Ta có:

B= 1/11+1/12+1/13+1/14+1/15+...+1/50

=(1/11+1/50)+(1/12+1/49)+...+(1/30+1/31)

=61/11.50+61/12.49+...+61/30.31

=61.(1/11.50+1/12.49+...+1/30.31)

Mình xin lỗi chỉ làm được đến đây vì dạng tính B mình không tốt lắm ◕◡◕

\(B=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{50}\right)>\left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)=> \(B>\frac{20}{30}+\frac{20}{50}=\frac{2}{3}+\frac{2}{5}=\frac{16}{15}>1\)

mà \(A=\frac{99}{100}

ssssssssss

2020.2022=(2021−1)(2021+1)=20212−1<\(2021^2\)

\(\Rightarrow2020.2021.2022< 2021^2.2021=2021^3\)