\(\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{n\times\left(n+1\right)}=\frac{49}{100}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{100}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{49}{100}\)

\(\Rightarrow\frac{n+1-2}{2\left(n+1\right)}=\frac{49}{100}\)

\(\Rightarrow\frac{n-1}{2n+2}=\frac{49}{100}\)

\(\Rightarrow100\left(n-1\right)=49\left(2n+2\right)\)

\(\Rightarrow100n-100=98n+98\)

\(\Rightarrow2n=198\)

=> n = 99

Vậy n =  99

\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+....+\(\frac{1}{n}\)-\(\frac{1}{n+1}\)=\(\frac{49}{100}\)

\(\frac{1}{2}\)-\(\frac{1}{n+1}\)=\(\frac{49}{100}\)

         \(\frac{1}{n+1}\)=\(\frac{1}{2}\)-\(\frac{49}{100}\)

          \(\frac{1}{n+1}\)=\(\frac{1}{100}\)

=> n+1=100

        n=100-1

       n=99

\(\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{nx\left(n+1\right)}=\frac{49}{100}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{100}\)

\(\frac{1}{2}-\frac{1}{n+1}=\frac{49}{100}\)

\(\frac{1}{n+1}=\frac{1}{2}-\frac{49}{100}\)

\(\frac{1}{n+1}=\frac{1}{100}\)

n + 1 = 100

n = 100 - 1

n = 99

Bn k cho mik nhé!

\(\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{a\times\left(a+1\right)}=\frac{49}{100}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{a}-\frac{1}{\left(a+1\right)}=\frac{49}{100}\)

\(\frac{1}{2}-\frac{1}{a+1}=\frac{49}{100}\)

\(\frac{50}{100}-\frac{1}{a+1}=\frac{49}{100}\)

\(\frac{1}{a+1}=\frac{50}{100}-\frac{49}{100}\)

\(\frac{1}{a+1}=\frac{1}{100}\)

\(\Rightarrow a+1=100\)

\(\Rightarrow a=100-1\)

\(a=99\)

Số a chính là: 99

Chúc bạn may mắn......mình chính là Đào Minh Tiến!

Dạ thưa Vũ a = 99

=1/2-1/3+1/3-1/4+.......+1/a-1/a+1=49/100

1/2-1/a+1=49/100

1/a+1 = 1/2-49/100

1/a+1=1/100

a+1=100

a=99

=1/2-1/3+1/3-1/4+.......+1/a-1/a+1=49/100

1/2-1/a+1=49/100

1/a+1 = 1/2-49/100

1/a+1=1/100

a+1=100

a=99

y=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

y=\(1-\frac{1}{6}\)

y=\(\frac{5}{6}\)

\(\Rightarrow y=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(\Rightarrow y=1-\frac{1}{6}=\frac{5}{6}\)

Vậy \(y=\frac{5}{6}\)

Nhưng hãy cho mình biết làm thế nào để làm ảnh đại diện 

trả lời xong rồi mình sẽ giải giúp bạn

(làm thế nào để đưa câu hỏi của mình lên online math)

..................................................................

\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{8x9}\)

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

=\(1-\frac{1}{9}\)

=\(\frac{8}{9}\)

OK XONG NHỚ CHO MIK NHA

\(\frac{1}{1\times2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+.......+\frac{1}{7x8}+\)\(\frac{1}{8x9}\)

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

=1-\(\frac{1}{9}\)

=\(\frac{8}{9}\)

\(\frac{1}{1\times2}+........+\frac{1}{8\times9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}=\frac{9}{10}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}\)

\(=\frac{2}{5}\)

\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{9x10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{9x10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}\)

\(=\frac{2}{5}\)