125 125 x 127 - 127 127 x 125

⇒ Dấu cách là dấu gì hả bạn?

a, (1 + 3 + 5 + 7+ ... + 2003 + 2005) x (125125 x 127 - 127127 x125)

= (1 + 3 + 5 +...+2003 + 2005) x (125 x 1001 x 127 - 127x1001 x 125)

= (1 + 3 + 5 +...+ 2003 + 2005) x 0

= 0 

b; 1000 x 0,04 - 100 : 2,5 + 10 x 0,4

= 100 x 0,4 - 100 x 0,4 + 10 x 0,4

= 0 + 4

= 4 

\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)

   \(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)

   \(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{1}{2}\left(\frac{1}{99}-\frac{1}{101}\right)\)

   \(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

   \(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)

   \(=\frac{1}{2}.\frac{98}{303}\)

   \(=\frac{49}{303}\)

Dấu chấm(.) ở cấp hai là dấu nhân (x)

\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{399}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)

\(=\frac{1}{2}.\left(\frac{7}{21}-\frac{1}{21}\right)\)

\(=\frac{1}{2}.\frac{6}{21}\)

\(=\frac{1}{7}\)

Chúc bạn học tốt !!! 

\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{399}\)

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)

\(2A=\frac{1}{3}-\frac{1}{21}=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}=\frac{2}{7}\)

\(A=\frac{2}{7}:2=\frac{2}{7}.\frac{1}{2}=\frac{2}{14}=\frac{1}{7}\)

Trả lời

Đặt A=\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{399}\)

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{19\cdot21}\)

\(2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{19\cdot21}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)

\(2A=\frac{1}{3}-\frac{1}{21}\)

\(2A=\frac{6}{21}\Rightarrow A=\frac{6}{21}:2\Rightarrow A=\frac{1}{7}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-...-\frac{1}{11.13}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(=1-\frac{1}{10}-\frac{1}{2}.\left(1-\frac{1}{13}\right)=\frac{9}{10}-\frac{6}{13}=\frac{57}{130}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{90}-\frac{1}{3}-\frac{1}{15}-.....-\frac{1}{143}\)

\(=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{143}\right)\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{9.10}\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{11.13}\right)\)

\(=\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)-\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)\(=\left(\frac{1}{1}-\frac{1}{10}\right)-\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{13}\right)=\frac{9}{10}-\frac{6}{13}=\frac{117}{130}-\frac{78}{130}=\frac{39}{130}=\frac{3}{10}\)

 E=4/3+4/15+4/35+4/63+...+4/399+4/483

=4/1.3+4/3.5+4/5.7+4/7.9+...+4/19.21+4/21.23

=2(2/1.3+2/3.5+2/5.7+...+2/19.21+2/21.23)

=2(1/1-1/3+1/3-1/5+1/5-1/7+...+1/19-1/21+1/21-1/23)

=2(1/1-1/23)

=2(23/23-1/23)

=2.22/23

=44/23

`1/15+1/35+1/63+1/99+1/143`

`=1/[3.5]+1/[5.7]+1/[7.9]+1/[9.11]+1/[11.13]`

`=1/2(2/[3.5]+2/[5.7]+2/[7.9]+2/[9.11]+2/[11.13])`

`=1/2.(1/3-1/5+1/5-1/7+...+1/11-1/13)`

`=1/2.(1/3-1/13)`

`=1/2 . 10/39`

`=5/39`

=\(\frac{1}{3\cdot5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{11\cdot13}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}.\frac{10}{13}=\frac{5}{13}\) 

B=1/15+1/35+1/63+1/99+1/143

B=1/3.5+1/5.7+1/7.9+1/9.11+11.13    (khoảng cách từ 3-5;5-7;7-9;9-11;11-13 la 2)

Suy ra B=1/2(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13) (Ta gop -1/5+1/5;-1/7+1/7;-1/9+1/9;-1/11+1/11 bang 0)

B=1/2(1/3-1/43)=1/2.40/129=20/129

\(A=\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{21.23}\)

A=\(\frac{1}{3}\left(1-\frac{1}{3}+\frac{1}{3}-.......-\frac{1}{23}\right)\)=\(\frac{1}{3}.\frac{22}{23}=\frac{22}{69}\)

hok t

tl lại

\(A=\frac{1}{1.3}+....\frac{1}{21.23}\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+.....+\frac{1}{21}-\frac{1}{23}\right)\)

\(A=\frac{1}{2}.\frac{22}{23}=\frac{11}{23}\)

k t nha

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{483}\)

\(=\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{21\times23}\)

\(=\frac{1}{2}\times[2\times\left(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{21\times23}\right)]\)

\(=\frac{1}{2}\times\left(\frac{2}{1\times3}\times\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{21\times23}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{21}-\frac{1}{23}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{23}\right)\)

\(=\frac{1}{2}\times\frac{22}{23}\)

\(=\frac{11}{23}\)

Nhớ h cho mk nha