\(=\left(1999\times1998+1998\times1997\right)\times\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times\left(\dfrac{4}{3}-\dfrac{4}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times0\)

\(=0\)

(1999×1998+1998+1997)×(1/1+1/2:3/2-4/3)

(1999×1998+1998+1997)×(1/1+1/2×2/3-4/3)

1999×1998+1998+1997)×(1/1+1/3-4/3)

(1999×1998+1998+1997)×(4/3-4/3)

(1999×1998+1998+1997)×0

0

bằng 0 nha bn

\(\left(1999.1998+1998+1997\right).\left(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\right)\)

\(=\left(1999.1998+1998+1997\right).\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)

\(=\left(1999.1998+1998+1997\right).\left(1+\frac{1}{3}-\frac{4}{3}\right)\)

\(=\left(1999.1998+1998+1997\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)

\(=\left(1999.1998+1998+1997\right).0=0.\)

( 1999 x 1998 + 1998 x 1997 ) x \(\left(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\right)\)

= ( 1999 x 1998 + 1998 x 1997 ) x  \(\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)

= ( 1999 x 1998 + 1998 x 1997 ) x  \(\left(1+\frac{1}{3}-\frac{4}{3}\right)\)

= ( 1999 x 1998 + 1998 x 1997 ) x 0

= 0

=19

0

\(D=\dfrac{1}{2000.1999}-\dfrac{1}{1999.1998}-\dfrac{1}{1998.1997}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(D=\dfrac{1}{1999.2000}-\left(\dfrac{1}{1998.1999}+\dfrac{1}{1997.1998}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)\(D=\dfrac{1}{1999.2000}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{1997.1998}+\dfrac{1}{1998.1999}+\dfrac{1}{1999.2000}\right)\)

\(D=\dfrac{1}{1999.2000}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{1997}-\dfrac{1}{1998}+\dfrac{1}{1998}-\dfrac{1}{1999}+\dfrac{1}{1999}-\dfrac{1}{2000}\right)\)\(D=\dfrac{1}{1999.2000}-\dfrac{1999}{2000}\)

a. ( 1999 x 1998 + 1998 x 1997 ) x ( 1 + 1/3 - 11/3 )

= ( 1999 x 1998 + 1998 x 1997 ) x ( 11/3 (Hỗn số) - 11/3)

= ( 1999 x 1998 + 1998 x 1997) x 0
= 0

b. 0,1 + 0,2 + 0,3 + 0,4 +......+ 0,19:

Khoảng cách giữa các số là : 0,2 - 0,1 = 0,1 ; 0,3 - 0,2 = 0,1 ; ....

Có số số hạng là : ( 0,19 - 0,1 ) : 0,1 + 1 = 19 (Mk đếm đó vì tính theo công thức thì ra 1,9 số hạng, số hạng không bao giờ dư nên mk đếm lun. Hì hì ^^)

Tổng là: ( 0,19 + 0,1 ) x 19 : 2 = 2,755.

                                                                                                                                    Đ/S : a) 0

                                                                                                                                             b) 2,755

Phần b Dao Minh Phuong làm sai rồi

a)(1999*1998+1998*1997)*(1+1/2:1 1/2-1 1/3)

=(1999*1998+1998*1997)*(1+1/2:3/2-4/3)

=(1999*1998+1998*1997)*(1+1/3-4/3)

=(1999*1998+1998*1997)*(4/3-4/3)

=(1999*1998+1998*1997)*0

=0

b)0.1+0.2+0.3+0.4+...+0.19

ta chia dãy số ra làm 2 dãy:

dãy 1:0.1+0.2+0..3+0.4+0.5+0.6+0.7+0.8+0.9

=(0.1+0.9)+(0.2+0.8)+(0.3+0.7)+(0.4+0.6)+0.5

=1           +       1     +    1       +      1      +   0.5

=4     +   0.5

=4.5

dãy 2:0.10+0.11+0.12+01.3+01.4+0.15+0.16+0.17+0.18+0.19

=(0.11+0.19)+(0.12+0.18)+.(0.13+0.17)+(0.14+0.16)+0.10+0.15

=0.3+0.3+0.3+0.3+0.10+0.15

=1.2+0.10+0.15

=1.45

Tổng của dãy số đó là :4.5+1.45=5.95