A=4000x4000

B=3998x4002

= (4000-2)x(4000+2)

= 4000x4000+4000x2-2x4000-2x2

= 4000x4000+8000-8000-4

= 4000x4000-4

Ta thấy:

A= 4000x4000

B= 4000x4000-4

=> A>B.

Ủng hộ nhé.

\(A=4000.4000\)\(=\left(3998+2\right).4000=3998.4000+8000\)

\(B=3998.4002=3998.\left(4000+2\right)\)\(=3998.4000+7996\)

Vì  \(8000>7996\) nên \(A>B\)

Nhớ nha các bạn

A=[(3999/2+1)+(3998/3+1)+...+(1/4000+1)+1]/(1/2+1/3+...+1/4001)

A=(4001/2+4001/3+...+4001/4001)/(1/2+1/3+...+1/4001)

A=[4001(1/2+1/3+...+1/4001)]/(1/2+1/3+...+1/4001)

A=4001

Vậy A=4001

! ) A = (3999 /2 +1 ) + ( 3998/ 3 + 1 ) + ( 3997 / 4 + 1 ) +...+ ( 1/ 4000 + 1 ) + 1

 (Ta lấy 4000/1 = 4000  rải đều  1, 1 ,1 cho 3999 phân số  và dư lại 1  = 4001/4001 )

= 4001 /2 + 4001 / 3 + 4001 /4 + ...+ 4001 /4000 + 4001 / 4001

= 4001 ( 1/2 + 1/3 + 1/4 +..+ 1/ 4001 )  vay A: B = 4001

Đặt A=\(\frac{4000}{1}+\frac{3999}{2}+\frac{3998}{3}+........+\frac{1}{4000}\)

A=\(1+\left(1+\frac{3999}{2}\right)+\left(1+\frac{3998}{3}\right)+........+\left(1+\frac{1}{4000}\right)\)

A=\(\frac{4001}{4001}+\frac{4001}{2}+\frac{4001}{3}+...........+\frac{4001}{4000}\)

A=\(4001.\left(\frac{1}{2}+\frac{1}{3}+........+\frac{1}{4000}+\frac{1}{4001}\right)\)

=>\(y=\frac{4001.\left(\frac{1}{2}+\frac{1}{3}+........+\frac{1}{4001}\right)}{\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{4001}}\)

=>\(y=4001\)

ai biết ko 1phần3+5phần8-7phần12

X x 2 = 3998

X = 3998 : 2

X = 1999

X x 2 = 3998

X = 3998 : 2

X = 1999