a, \(\frac{1}{2}.B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

      \(\frac{1}{2}.B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

         \(\frac{1}{2}.B=1-\frac{1}{101}=\frac{100}{101}\)

                  \(B=\frac{100}{101}.2=\frac{200}{101}\)

b, \(\frac{4}{5}.C=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{101.105}\)

      \(\frac{4}{5}.C=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)

          \(\frac{4}{5}.C=1-\frac{1}{105}=\frac{104}{105}\)

                 \(C=\frac{104}{105}.\frac{5}{4}=\frac{26}{21}\)

\(B=\frac{2}{2}\cdot\left(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+....+\frac{4}{99\cdot101}\right)\)

\(=\frac{4}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\)

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

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

\(=2\cdot\frac{100}{101}\)

\(=1\frac{99}{101}\)

a) \(B=\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{99\cdot101}\)

\(\Rightarrow\frac{2}{4}B=\frac{2}{4}\left(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{99\cdot101}\right)\)

\(\Leftrightarrow\frac{2}{4}B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{99\cdot101}\)

\(\Leftrightarrow\frac{2}{4}B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{101}\)

\(\Leftrightarrow\frac{2}{4}B=1-\frac{1}{101}=\frac{100}{101}\)

\(\Leftrightarrow B=\frac{100}{101}:\frac{2}{4}=\frac{100\cdot4}{101\cdot2}=\frac{200}{101}\)

2+12345678-5=

cậu nhâ n cả 2 vế của C với 4 / 5 ý ra lu n ak

\(C=\frac{5}{1.5}+\frac{5}{5.9}+\frac{5}{9.13}+...+\frac{5}{101.105}\)

\(C=5.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{101.105}\right)\)

\(C=5.\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{101}-\frac{1}{105}\right)\)

\(C=\frac{5}{4}.\left(1-\frac{1}{105}\right)\)

\(C=\frac{5}{4}.\frac{104}{105}\)

\(C=\frac{26}{21}\)

\(C=\frac{5}{1.5}+\frac{5}{5.9}+\frac{5}{9.13}+...+\frac{5}{101.105}\)

\(\Rightarrow\frac{4}{5}C=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{101.105}\)

\(\Rightarrow\frac{4}{5}C=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{101}-\frac{1}{105}\)

\(\Rightarrow\frac{4}{5}C=1-\frac{1}{105}\)

\(\Rightarrow\frac{4}{5}C=\frac{104}{105}\)

\(\Rightarrow C=\frac{104}{105}.\frac{5}{4}=\frac{26}{21}\)

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{3\left(x-1\right)\left(3x+3\right)}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{\left(3x-1\right)\left(3x+3\right)}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)(Vì 3x + 3 lớn hơn 3x - 1 là 4 đơn vị)

\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{x+1-1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{x}{3x+3}=\frac{3}{10}\)

\(\Rightarrow10x=3.\left(3x+3\right)\)

\(\Rightarrow10x=9x+9\)

\(\Rightarrow x=9\)

Vậy...

thanks

\(E=\frac{\frac{4}{3\cdot7}-\frac{4}{11.15}}{1-\frac{3}{7}-\frac{3}{11}+\frac{1}{5}}-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2006.2007}\right)\)

\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{11}+\frac{1}{15}}{\frac{192}{385}}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2006}-\frac{1}{2007}\right)\)

\(=\frac{\frac{64}{385}}{\frac{192}{385}}-\left(\frac{1}{3}-\frac{1}{2007}\right)\)

\(=\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{2007}\right)=\frac{1}{2007}\)

Vậy : \(E=\frac{1}{2007}\)

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}+0+0+0+0\)

\(=\frac{8}{27}\)

Ta có : \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}\)

\(=\frac{8}{27}\)

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{7-3}{3.7}+\frac{11-7}{7.11}+.....+\frac{27-23}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-....-\frac{1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3x+3}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)

Nên 3x + 3 = 30

3x = 30 - 3 = 27

x = 27 : 3 = 9 

4( 1/3.7 + 1/7.11 +...+1/(3x-1)(3x+3) = 3/10

4( 1/3 - 1/7 + 1/7 - 1/11 +...+ 1/(3x-1) - 1/(3x+3) =3/10

4(1 -  1/(3x+3))= 3/10

phần còn lại tự giải nghe x=-71/111

\(\Leftrightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)

\(\Rightarrow=\frac{1}{3}-\frac{1}{111}\)

\(=\frac{12}{37}\)

k nha

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)

\(=\frac{1}{3}-\frac{1}{111}\)

\(=\frac{108}{333}=\frac{12}{37}\)

ten gia mao nobita 

nobi k the gioi nhu the 

do gia :v