1/3.4+1/4.5+1/5.6+.....+1/x(x+1)=3/10

1/3-1/4+1/4-1/5+1/5-........-1/x+1/x-1/x+1=3/10

=>1/3-1/x+1=3/10

   1/x+1=3/10-1/3=1/30

=>x+1=30

   x=30-1

   x=29

Ta có :

\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)

=>\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)

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

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

=>\(\frac{1}{x+1}=\frac{1}{30}\)

=>\(x+1=30\)

=>\(x=30-1\)

=>\(x=29\)

Vậy \(x=29\)

\(2\)

CMR

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

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

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

\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)

mình biết làm mỗi ý thứ 2 thôi.

(1/3-1/4+1/4-1/5+1/5-.......+1/x.(x+1)=3/10

1/3-1/x+1=3/10

tự làm...

1/3.4+1/4.5+1/5.6+1/6.7+....+1/x(x+1)=3/10

<=> \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(x+1\right)x}=\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)

<=> \(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)=> x+1=30=>x=29

\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)

\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)

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

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

\(\frac{1}{x+1}=\frac{1}{30}\)

\(\Rightarrow x+1=30\)

\(x=30-1=29\)

Quá dễ:

=> 1/3 - 1/4 + 1/4 - 1/5 + ....+ 1/x - 1/x+1 = 3/10

=> 1/3 - 1/x+1 = 3/10

=> 1/x+1 = 1/3 - 3/10

Còn lại tự làm nhá!

<=> 1/3 - 1/(x+1) = 3/10 

<=> 1/(x+1) = 1/30

=> x+1 = 30 

<=> x= 29 

= > 1/3 - 1/4 +1/4 -1/5 + ..... + 1/x - 1/x+1 = 3/10

= > 1/3 - 1/x+1 = 3/10

= > 1/x+1         = 1/3 - 3/10

= > 1/x+1         =1/30 

= > x+1 = 30

= > x     = 29

Vậy x = 29

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{3}{10}\)

\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)

\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{n+1}=\dfrac{3}{10}\)

\(\Rightarrow\dfrac{1}{n+1}=\dfrac{1}{30}\)

\(\Rightarrow n+1=30\)

\(\Rightarrow n=29\)

Vậy n = 29.

\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + .....+\(\dfrac{1}{n.(n+1)}\) = \(\dfrac{3}{10}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\) +......+ \(\dfrac{1}{n}-\dfrac{1}{n+1}\) = \(\dfrac{3}{10}\)

\(\dfrac{1}{3}-\dfrac{1}{n+1}\) = \(\dfrac{3}{10}\)

         \(\dfrac{1}{n+1}\) = \(\dfrac{1}{3}-\dfrac{3}{10}\)

          \(\dfrac{1}{n+1}\) = \(\dfrac{1}{30}\)

           n + 1 = 30

           n = 30 - 1

           n = 29

Kết luận n = 29 là giá  trị thỏa mãn yêu cầu đề bài.

\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)

\(\dfrac{1}{3}-\dfrac{1}{n+1}=\dfrac{3}{10}\)

\(\dfrac{-1}{\left(n+1\right)}=\dfrac{-1}{30}\)

\(-n-1=-30\)

-n = -29

n = 29

nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

bài này có x mà ko có vế phải à

chịu, cô tui cho đề thế mà