a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}=\dfrac{1}{2}\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}\)

b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)

a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}\)

\(=\dfrac{n}{2n+1}\)

b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)

\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)

F=(3/1.8+3/8.15+...+3/106.113)-(25/50.55+25/55.60+...+25/95.100) (1)

Đặt:

A=(3/1.8+3/8.15+3/15.22+...+3/106.113)

=3/7.(1-1/8+1/8-1/15+...+1/106-1/113)

=3/7.(1-1/113)

=3/7.112/113

=336/791. (2)

B=25/50.55+25/55.60+...+25/95.100

=25/5.(1/50-1/55+1/55-1/60+...+1/95-1/100)

=5.(1/50-1/100)

=1/20 (3)

Thay (2),(3) vào (1) ta được:

F=336/791-1/20

=5929/6720.

Chỗ phức tạp là ở biểu thức trong ngoặc thôi

Ta có

\(\dfrac{1}{8}+\dfrac{1}{8\cdot15}+\dfrac{1}{15\cdot22}...+\dfrac{1}{43\cdot50}\)

\(=\dfrac{1}{8}\cdot\left[\dfrac{1}{7}\left(\dfrac{1}{8}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{22}+....+\dfrac{1}{43}-\dfrac{1}{50}\right)\right]\)

\(=\dfrac{1}{8}\cdot\left[\dfrac{1}{7}\left(\dfrac{1}{8}-\dfrac{1}{50}\right)\right]=\dfrac{1}{8}\cdot\dfrac{3}{200}=\dfrac{3}{1600}\)

Bài 1 :

\(\left(\frac{1}{8}+\frac{1}{8.15}+\frac{1}{15.22}+...+\frac{1}{43.50}\right)\frac{4-3-5-7-...-49}{217}\)

\(=\frac{1}{7}\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+...+\frac{1}{43}-\frac{1}{50}\right).\frac{5-\left(1+3+5+7+...+49\right)}{217}\)

\(=\frac{1}{7}\left(1-\frac{1}{50}\right).\frac{5-\left(12.50\right)+25}{217}\)

\(=\frac{1}{7}.\frac{49}{50}.\frac{5-625}{217}\)

\(=\frac{-2}{5}\)

Bài 2 :

\(B=\frac{x^2+17}{x^2+7}=\frac{\left(x^2+7\right)+10}{x^2+7}=1+\frac{10}{x^2+7}\)

Ta có : \(x^2\ge0\). Dấu '' = '' xảy ra khi :

\(x=0\Rightarrow x^2+7\ge7\)( 2 vế dương )

\(\Rightarrow\frac{10}{x^2+7}\le\frac{10}{7}\)

\(\Rightarrow1+\frac{10}{x^2+7}\le1+\frac{10}{7}\)

\(\Rightarrow B\le\frac{17}{7}\)

Dấu '' = '' xảy ra < = > x = 0

Vậy Max \(B=\frac{17}{7}\Leftrightarrow x=0\) 

\(C=\frac{49}{1.8}+\frac{49}{8.15}+\frac{49}{15.22}+...+\frac{49}{141.148}\)

\(\Rightarrow\frac{1}{7}C=\frac{7}{1.8}+\frac{7}{8.15}+\frac{7}{15.22}+...+\frac{7}{141.148}\)

\(=1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+...+\frac{1}{141}-\frac{1}{148}\)

\(=1-\frac{1}{148}=\frac{148}{148}-\frac{1}{148}=\frac{147}{148}\)

\(\Rightarrow C=\frac{147}{148}\div\frac{1}{7}=\frac{147}{148}.7=\frac{1029}{148}\)

Vậy \(C=\frac{1029}{148}.\)

=49(1/1.8 +1/8.15+..)

có

1/1.8= (1-1/8)

1/8.15=1/8-1/15

=> C=49*1/7*(1-1/8+1/8-1/15+...+1/141-1/148)

C=7*(1-1/148)=...

học tốt

Bài làm:

Ta có: \(C=\frac{49}{1.8}+\frac{49}{8.15}+\frac{49}{15.22}+...+\frac{49}{141.148}\)

\(C=7\left(\frac{8-1}{1.8}+\frac{15-8}{8.15}+\frac{22-15}{15.22}+...+\frac{148-141}{141.148}\right)\)

\(C=7\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+...+\frac{1}{141}-\frac{1}{148}\right)\)

\(C=7\left(1-\frac{1}{148}\right)\)

\(C=7.\frac{147}{148}=\frac{1029}{148}\)

(-7/15) . 5/8 . 15/-7 . (-32)

=( -7/15 . 15/-7 ) . [ 5/8 . (-32)]

= 1 . ( -20 )

= -20

trả lời -20