a) \(\frac{1}{n}\) - \(\frac{1}{n+1}\) = \(\frac{n+1}{n\left(n+1\right)}\) - \(\frac{n}{n\left(n+1\right)}\) = \(\frac{1}{n\left(n+1\right)}\) = \(\frac{1}{n}\) . \(\frac{1}{n+1}\) =>đpcm

b) A= \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\)+...+\(\frac{1}{8}\) - \(\frac{1}{9}\) +\(\frac{1}{9}\)

= \(\frac{1}{2}\) + \(\frac{1}{9}\)= \(\frac{11}{18}\)

Câu 1 :\(P=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{100}=\frac{1}{100}\)

like mình làm hết

a)\(\frac{1}{a}-\frac{1}{a+1}=\frac{a+1}{a\left(a+1\right)}-\frac{a}{\left(a+1\right)a}=\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}\)

b)\(\frac{1}{a}-\frac{1}{b}=\frac{b}{ab}-\frac{a}{ab}=\frac{b-a}{ab}\)

Không chép lại đề nhé

Ta có:

P=\(\frac{50-49}{49}+\frac{50-48}{48}+...+\frac{50-2}{2}+\frac{50-1}{1}\)

P=\(\frac{50}{49}-\frac{49}{49}+\frac{50}{48}-\frac{48}{48}+...+\frac{50}{2}-\frac{2}{2}+\frac{50}{1}-\frac{1}{1}\)

P=\(\left(\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\right)+\frac{50}{1}-\left(\frac{49}{49}+\frac{48}{48}+...+\frac{2}{2}+\frac{1}{1}\right)\)

P=\(50\cdot\left(\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)+50-49\)                 (chỗ này gộp nha)

P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{48}+\frac{1}{49}\right)+1\)

P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)+\frac{50}{50}\)

P=\(50\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)\)

=>P=50S

=>\(\frac{S}{P}=\frac{S}{50S}=\frac{1}{50}\)

Vừa nãy mình nói nhầm, Sorry.

Tích nha

Chào bạn, bạn hãy theo dõi bài giải của mình nhé!

Ta có : 

\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)

\(=>A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

\(=>A=\frac{3.4.5.....100}{2.3.4.....99}=\frac{100}{2}=50\)

Chúc bạn học tốt!

sao ban biet 3.4.5...10 = 100

1+1/3+1/6+...+1/x(x+1)=1/2003/2005

    1/3+1/6+...+1/x(x+1)=2003/2005

   1/2(1/3+1/6+..+1/x(x+1)=2003/4010

 1/6+1/12+...+1/x(x+1)=2003/4010

  1/2*3+1/3*4+...+1/x(x+1)=2003/4010

   1/2-1/3+1/3-1/4+...+1/x-1/x+1=2003/4010

 1/2-1/x+1=2003/4010

         1/x+1=1/2005

           x+1=2005

          x=2004

lộn

\(\dfrac{1}{k^2}<\dfrac{1}{k(k-1)}=\dfrac{1}{k-1}-\dfrac{1}{k}\)

Ap dung:

\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\ldots+\dfrac{1}{n^2}<1+\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\ldots+\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)=2-\dfrac{1}{n}<2\)

Ta có: \(\frac{1}{1^2}=\frac{1}{1\cdot1};\frac{1}{2^2}<\frac{1}{1\cdot2};...;\frac{1}{50^2}<\frac{1}{49\cdot50}\)

=>\(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}<1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}=1+1-\frac{1}{50}=2-\frac{1}{50}=1,98\)

hay A<1,98 mà 1,98<2 nên A<2

Vậy A<2

M=1+1/2^2+1/3^2+1/4^2+...+1/10^2>1+1/2*3+1/3*4+1/4^5+...+1/10*11

                             M>1+1/2-1/3+1/4-1/4+1/5-...-1/11

                            M>1+1/2-1/11

                              M>1+9/22

                               M>31/22

                                vì 31/22>4/3 nên M>4/3

khúc đằng trước hỉu j chết liền

lên olm đăg thử đi