Giải:

\(A=\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{98.99}+\dfrac{9}{99.100}\)

\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=9.\left(1-\dfrac{1}{100}\right)\)

\(A=9.\dfrac{99}{100}\)

\(A=\dfrac{891}{100}\)

A=9/1.2+9/2.3+9/3.4+.....+9/98.99+9/99.100

  =9.(1/1.2+1/2.3+1/3.4+....+1/98.99+1/99.100

 =9.(1/1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100)

 =9.(1/1-1/100)

 =9.99/100

 =891/100

CHÚC BẠN HỌC TỐT!

\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

\(=9.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=9.\left(1-\frac{1}{100}\right)\)

\(=9.\frac{99}{100}\)

\(=\frac{891}{100}\)

A=9/1.2+9/2.3+9/3.4+.....+9/98.99+9/99.100

  =9.(1/1.2+1/2.3+1/3.4+....+1/98.99+1/99.100

 =9.(1/1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100)

 =9.(1/1-1/100)

 =9.99/100

 =891/100

a) 100 - 99 + 98 -97 + 96 -95 +...+ 4-3 + 2

= (100 - 99) + (98 -97) + (96 - 95) +...+ (4-3) +2 (gồm 49 cặp và 1 số hạng)

= 1+1+1+....+1 +2

= 49 x 1 + 2 = 51

b) 100 - 5-...-5 - 5  (20 số 5)

= 100 - 20 x 5 = 0

c) 99 - 9 - 9 -... - 9 -9 (11 số 9)

=99 - 11 x 9 = 0

d) 2011 + 2011+2011+2011 - 2008 x 4

= 2011 x 4 - 2008 x 4

= 4 x (2011 - 2008)

= 4 x 3

=12

Giải:

\(A=\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{98.99}+\dfrac{9}{99.100}\)

\(\Leftrightarrow A=9\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(\Leftrightarrow A=9\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=9\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=9.\dfrac{99}{100}\)

\(\Leftrightarrow A=\dfrac{891}{100}\)

Vậy ...

Số các số hạng là:

(2000 - 100) : 1 + 1 = 1901

Tổng là:

(2000 + 100) x 1901 : 2 = 1996050

Đáp số : 1996050

= [(2000-100)+1]: 2 x (2000+100)= 1996050

Tổng số các số hạng là :

( 2 000 - 100 ) : 1 + 1 = 1 901 ( số hạng )

Tổng của dãy số trên là :

( 2 000 + 100 ) x 1 901 : 2 = 1 996 050

Đáp số : 1 996 050

ủng hộ mk nha các bn ^-^

\(\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{99.100}\)

=\(9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

=\(9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

=\(9.\left(\frac{1}{1}-\frac{1}{100}\right)\)

=\(9.\frac{99}{100}\)

=\(\frac{891}{100}\)

\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

\(A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=9\left(1-\frac{1}{100}\right)\)

\(A=9\times\frac{99}{100}\)

\(A=\frac{891}{100}\) hoặc 8,91

\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

A=\(9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=9.\left(1-\frac{1}{100}\right)=\frac{891}{100}\)