a) thấy dấu cộng ở trước số 6 thành dấu trừ

b) =  2/ 2 + 2/ 6 + 2/ 12 + 2/ 20 + 2/ 30 + 2/ 42 + 2/ 56 + 2/ 72 + 2/ 90

= 2x ( 1/ 1x2 + 1 / 2x3 + 1/ 3x4 + 1/ 4x5 + 1/ 5x6 + 1/ 6x7 + 1/ 7x8 + 1/ 8x9 + 1/ 9x10 )

= 2x ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +1/5 - 1/6 +.. + 1/8- 1/9  + 1/9 - 1/10 )

=2 x( 1 - 1/10 ) 

=2 x 9/10 = 18/10 = 9 / 5

không nhé

(2x+1)(4x^2-xy+1)-(8x^3-1)

= ((2x)^3 -1) - ( 8x^3 - 1 ) = 0

Vậy là không phụ thuộc vào biến nhé bạn

Bài 1:

\(A=\frac{5}{3.6}+\frac{5}{6.9}+....+\frac{5}{96.99}\)

\(\Rightarrow\frac{3}{5}A=\frac{3}{3.6}+\frac{3}{6.9}+....+\frac{3}{96.99}\)

\(\Rightarrow\frac{3}{5}A=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

\(\Rightarrow A=\frac{32}{99}\div\frac{3}{5}=\frac{160}{297}\)

Bái 2:

\(B=\frac{2}{3.7}+\frac{2}{7.11}+...+\frac{2}{99.103}\)

\(\Rightarrow2B=\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{99.103}\)

\(\Rightarrow2B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{99}-\frac{1}{103}\)

\(=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)

\(\Rightarrow B=\frac{100}{309}\div2=\frac{50}{309}\)

Bài 1:

Ta có:

\(\frac{5}{n.\left(n+3\right)}=\frac{5}{3}.\frac{3}{n.\left(n+3\right)}=\frac{5}{3}.\frac{\left(n+3\right)-n}{n.\left(n+3\right)}=\frac{5}{3}.\left[\frac{n+3}{n.\left(n+3\right)}-\frac{n}{n\left(n+3\right)}\right]\)\(=\frac{5}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)\)

\(\frac{5}{3.6}+\frac{5}{6.9}+\frac{5}{9.12}+...+\frac{5}{96.99}=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\right)\)

\(=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{160}{297}\)

ta có 
1/2 P=1/2(1-1/10-1/15-1/3-1/28-1/6-1/21)
=1/2-(1/6+1/12+1/20+1/30+1/42+1/56)
=1/2-(1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8)
=1/2-(1/2-1/8)
=1/8
suy ra P=1/4

ta có 
1/2 P=1/2(1-1/10-1/15-1/3-1/28-1/6-1/21)
=1/2-(1/6+1/12+1/20+1/30+1/42+1/56)
=1/2-(1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8)
=1/2-(1/2-1/8)
=1/8
suy ra P=1/4

A=/x-2/+3x+2