Đặt \(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{17x19}\)

=>\(2xA=2x\left(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{17x19}\right)\)

=>\(2xA=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{17x19}\)

=>\(2xA=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)

=>\(2xA=1-\frac{1}{19}=\frac{18}{19}\)

=>\(A=\frac{18}{19}:2=\frac{9}{19}\)

(\(\frac{1}{1}-\frac{1}{3}\left(\right)+\left(\right)\frac{1}{3}-\frac{1}{5}\left(\right)+\left(\right)\frac{1}{5}-\frac{1}{7}\left(\right)+....+\left(\right)\frac{1}{17}-\frac{1}{19}\left(\right)\)\(\frac{1}{19}\)

\(\frac{1}{1}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+....+\left(\frac{1}{17}-\frac{1}{17}\right)-\frac{1}{19}\)

\(\frac{1}{1}-\frac{1}{19}=\frac{18}{19}\)

=2x(\(\frac{1}{1\times3}+\)\(\frac{1}{3\times5}+\)\(\frac{1}{5\times7}+\)..........+\(\frac{1}{17\times19}\))

=\(\frac{2}{1\times3}+\)\(\frac{2}{3\times5}+\)\(\frac{2}{5\times7}+\)............+\(\frac{2}{17\times19}\)

=\(\frac{1}{1}-\)\(\frac{1}{3}+\)\(\frac{1}{3}-\)\(\frac{1}{5}+\)\(\frac{1}{5}-\frac{1}{7}\)\(+\)..........\(+\)\(\frac{1}{17}-\frac{1}{19}\)

=\(\frac{1}{1}-\frac{1}{19}\)

=\(\frac{19}{19}-\frac{1}{19}\)

=\(\frac{18}{19}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)

\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{93}\)

\(\Rightarrow2x+3=93\)

\(\Rightarrow2x=90\)

\(\Rightarrow x=45\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{96}\)

\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{96}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{5}{16}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{5}{16}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{5}{16}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{5}{16}\)

\(\frac{1}{2x+3}=\frac{1}{48}\)

=> 2x + 3 = 48

=> 2x = 48 - 3

=> 2x = 45

=> x = 45/2

1=3/3=4/4=5/5=...

=> 1+1/1*3=3/1*3=1/1

=> 1+1/2*4=4/2*4=1/2

=>...

Bieu thuc se con lai la 1*1/2*1/3*1/4*1/5

Vay A=1/120

(4/1*3+4/3*5+4/5*7+4/7*9)*10-x=0

=4*2/1*3+4*2/3*5+4*2/5*7+4*2/7*9

=1/1+1/3+1/5+1/7+1/9

=1/1-1/9

=8/9

8/9*10-x=0

89-x=0

x=89-0

x=89

a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)

\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)

\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)

chết phần a quên nhân vs 1/3

b)\(VT=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)

\(=\frac{5}{4}\left[\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right]\)

\(=\frac{5}{4}\cdot\left[\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right]\)

\(=\frac{5}{4}\cdot\left[\frac{1}{3}-\frac{1}{4n+3}\right]=\frac{5}{4}\cdot\left[\frac{4n+3}{3\left(4n+3\right)}-\frac{3}{3\left(4n+3\right)}\right]\)

\(=\frac{5}{4}\cdot\left[\frac{4n+3-3}{12n+9}\right]\)\(=\frac{5}{4}\cdot\frac{4n}{12n+9}=\frac{5n}{12n+9}\)

\(\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\times y=\frac{2}{3}\)

\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\times y=\frac{2}{3}\)

\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{11}\right)\times y=\frac{2}{3}\)

\(\frac{1}{2}\times\frac{10}{11}\times y=\frac{2}{3}\)

\(\frac{5}{11}\times y=\frac{2}{3}\) => \(y=\frac{2}{3}:\frac{5}{11}=\frac{2}{3}\times\frac{11}{5}=\frac{22}{15}\)