nhấn đúng 0 ra đáp án nha

2/3+2/15+2/35+...+2/n x (n+2)=322/323

=2/1.3+2.3.5+2/5.7+...+2/n x(n+2)=322/323

=                                         n+2 =323

                                             n=323-2

                                              n=321    

2/3+2/15+2/35+...+2/n x (n+2) = 322/323

2/1x3+2/3x5+2/5x7 +...+ 2/nx(n+2) = 322/323

1-1/3+1/3-1/5+1/5-1/7+...+1/n-1/(n+2) = 322/323

1-1/n+2 = 322/323

1/n+2 = 1-322/323

1/n+2 = 1/323

=> n+2 = 323

n = 323 - 2 = 321

cái này tính cái gì thế

ko hiểu

a)\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{2}{15}\)

b)\(M=1+3+3^2+...+3^{25}=\frac{3^{26}-1}{3-1}=\frac{3^{26}-1}{2}

bạn đọc lại đề bài b) đi

a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) 

vì  ( 125125 x 127 - 127127 x 125 ) =[125125 x (125+2)] - 127127 x 125 ) =>125125 x (125+2)=125.125125+125125.2=125125.125+250250=125125.125+125.2002=125.(125125+2002)=125.127127

=> ( 125125 x 127 - 127127 x 125 )=127127.125-127127.125=0

=>  (1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) =0

a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) 

= ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 )  x 0

= 0

b, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)

= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)

= \(\frac{1}{3}\)- \(\frac{1}{15}\)

= \(\frac{4}{15}\)

\(5-\dfrac{2}{3}-\dfrac{14}{15}+\dfrac{1}{35}-\dfrac{62}{63}-\dfrac{98}{99}-\dfrac{142}{143}\)

\(=5-\left(1-\dfrac{1}{3}\right)-\left(1-\dfrac{1}{15}\right)+\dfrac{1}{35}-\left(1-\dfrac{1}{63}\right)-\left(1-\dfrac{1}{99}\right)-\left(1-\dfrac{1}{143}\right)\)

\(=5-1+\dfrac{1}{1\cdot3}-1+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}-1+\dfrac{1}{7\cdot9}-1+\dfrac{1}{9\cdot11}-1+\dfrac{1}{11\cdot13}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

\(=1-\dfrac{1}{13}=\dfrac{12}{13}\)

=> 2/1x3 +2/3x5+2/5x7+2/7x9+...+2/nx(n+2)

=>1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9+...+1/n-1/n+2

=>1-1/n+2=100/101

1/n+2=1-100/101

1/n+2=1/101

=>n+2=101

=>n=101-2

=>n=99

1: =(195-21):2+108=174:2+108=87+108=195

2: \(=16\left(444446-444444\right)=16\cdot2=32\)

3: \(=\dfrac{5^{3015}}{5^{3013}}-\dfrac{5^{3013}}{5^{3013}}=25-1=24\)

4: \(=125-\left[65-4^3\right]=125-1=124\)

6: \(=8210\cdot43:15\cdot86=2024038.667\)