(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

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

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

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

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

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

\(\Rightarrow x+1=30\)

\(x=30-1=29\)

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bài này có x mà ko có vế phải à

chịu, cô tui cho đề thế mà

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 \(x=29\)

Theo đề suy ra

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

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

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

=>x+1=30

=>x=29

6,45

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\\ =\dfrac{1}{2}-\dfrac{1}{7}\\ =\dfrac{5}{14}\)

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}-\dfrac{1}{5\cdot6}-\dfrac{1}{6\cdot7}-\dfrac{1}{7\cdot8}-\dfrac{1}{8\cdot9}\)

`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

`=`\(\dfrac{1}{3}-\left(\dfrac{1}{2}-\dfrac{1}{9}\right)\)

`=`\(\dfrac{1}{3}-\dfrac{7}{18}=-\dfrac{1}{18}\)

ta phân tích thành

\(\frac{1}{1\cdot2}\)x\(\frac{2\cdot2}{2\cdot3}\)x\(\frac{3\cdot3}{3\cdot4}\)x......x\(\frac{5\cdot5}{5\cdot6}\)x\(\frac{6\cdot6}{6\cdot7}\)

Tử nhân tử mẫu nhân mẫu ta có

1x2x2x3x3x.........x5x5x6x6

1x2x2x3x3x4x....x5x6x6x7

Rút gọn ta có

\(\frac{1}{7}\)

Vậy A=\(\frac{1}{7}\)