\(B=\frac{2^3}{3.5}+\frac{2^3}{5.7}+....+\frac{2^3}{101.103}\)

\(\Rightarrow\frac{1}{2^2}.B=\frac{2}{3.5}+\frac{2}{4.7}+....+\frac{2}{101.103}\)

\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\)

\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)

\(\Rightarrow B=\frac{100}{309}:\frac{1}{4}=\frac{400}{309}\)

\(=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{103}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{103}\right)\)

\(=4\cdot\frac{100}{309}=\frac{400}{309}\)

a)Ta có:

\(A=4\frac{25}{16}+25\left(\frac{9}{16}:\frac{125}{64}\right):\frac{-27}{8}\)

\(\Rightarrow A=\frac{89}{16}+25.\frac{36}{125}:\frac{-27}{8}\)

\(\Rightarrow A=\frac{89}{16}+\frac{36}{5}:\frac{-27}{8}\)

\(\Rightarrow A=\frac{89}{16}+\frac{-32}{15}\)

\(\Rightarrow A=\frac{823}{240}\)

Vậy A=.....

b)Ta có:

\(C=\frac{2^3}{3.5}+\frac{2^3}{5.7}+\frac{2^3}{7.9}+...+\frac{2^3}{101.103}\)

\(\Rightarrow C=\frac{2^2.2}{3.5}+\frac{2^2.2}{5.7}+\frac{2^2.2}{7.9}+...+\frac{2^2.2}{101.103}\)

\(\Rightarrow C=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)

\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{101}-\frac{1}{103}\right)\)

\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{103}\right)\)

\(\Rightarrow C=4.\frac{100}{309}\)

\(\Rightarrow C=\frac{400}{309}\)

Vậy C=.....

B, C=2^3/3.5 + 2^3/5.7+......+2^3/101.103

C= 2^3(1/3-1/5+1/5-1/7+....+1/101-1/103)

C=8(1/3-1/103)

C=8.100/309

C=800/309

VẬY C= 800/309

a.  

\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)

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

b.

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)

mk đầu tiên nha bạn

\(p=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2009.2011}\)
\(p=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2009}-\frac{1}{2011}\)
\(p=\frac{1}{3}-\frac{1}{2011}\)
\(p=\frac{2011}{6033}-\frac{3}{6033}\)
\(p=\frac{2008}{6033}\)

Ta có S=2/3+2/3.5+2/5.7+2/7.9+...+2/97.99 
           =2/3+1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
           =2/3+1/3+(1/5-1/5)+(1/7-1/7)+...+(1/97-1/97)+1/99
           =1+0+0+0+...+0+1/99
           =1+1/99
           =100/99
Mà 100/99>1.Suy ra S>1
   Vậy S>1
            

S=1-1/3 + 1/3 - 1/5 + ... + 1/97 - 1/99

=1 - 1/99 => S<1

a: \(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}=\dfrac{1}{3}-\dfrac{1}{203}=\dfrac{200}{609}\)

b: \(B=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{73}-\dfrac{1}{76}\)

\(=\dfrac{1}{4}-\dfrac{1}{76}=\dfrac{18}{76}=\dfrac{9}{38}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)

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

~ Hok tốt ~

\(\)

Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99

Tui hk bít nữa