`@` `\text {Ans}`

`\downarrow`

`a)`

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

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

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

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

`=`\(\dfrac{2}{9}\)

Vậy, \(A=\dfrac{2}{9}\)

`b)`

\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)

Vậy, \(B=\dfrac{4}{25}\)

`c)`

\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)

`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)

Vậy, \(C=\dfrac{99}{100}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)\(\frac{1}{25}\)

\(A=\frac{1}{5}-\frac{1}{25}\)

\(A=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

\(A=\frac{1}{5}-\frac{1}{25}\)

\(A=\frac{4}{25}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+.......+\frac{1}{24}-\frac{1}{25}\)

\(A=\frac{1}{5}+\left(-\frac{1}{6}+\frac{1}{6}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+\left(-\frac{1}{8}+\frac{1}{8}\right)+.......+\left(-\frac{1}{24}+\frac{1}{24}\right)-\frac{1}{25}\)       

\(A=\frac{1}{5}-\frac{1}{25}=\frac{4}{25}\)

\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+.........+\frac{1}{24.25}\)

= \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-........-\frac{1}{24}+\frac{1}{24}-\frac{1}{25}\)

= \(\frac{1}{5}-\frac{1}{25}\)

= \(\frac{5}{25}-\frac{1}{25}\)

= \(\frac{4}{25}\)

Ta có : 

\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(=\)\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\)\(\frac{1}{5}-\frac{1}{25}\)

\(=\)\(\frac{5}{25}-\frac{1}{25}\)

\(=\)\(\frac{4}{25}\)

Vậy \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}=\frac{4}{25}\)

Chúc bạn học tốt ~

\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(=1\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\right)\)

\(=1\left(\frac{1}{5}-\frac{1}{25}\right)\)

\(=\frac{4}{25}\)

A=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

A=1/3-1/9

A=2/9

các câu 2;3 còn lại giống câu 1 bạn nhé

bạn thay số vào rồi làm tương tự

A=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

A=1/3-1/9

A=2/9.

1/5.6 + 1/6.7 + 1/7.8 +...+ 1/24.25

=1/5 - 1/6 + 1/6-1/7 +1/7-1/8 + ... + 1/24-1/25

=> Kết quả là: 1/5 - 1/25 = 4/25

b) 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9+...+ 2/99.101

=2/1-2/3 + 2/3-2/5 + 2/5-2/7 + 2/7-2/9 + ... + 2/99-2/101

=> kết quả là 2/1 - 2/101 =200/101

a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

=\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

=\(\frac{1}{5}-\frac{1}{25}\)

=\(\frac{4}{25}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

=\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)

=\(2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

=\(2.\left(\frac{1}{1}-\frac{1}{101}\right)\)

=\(2.\frac{100}{101}\)

=\(\frac{200}{101}\)

a)\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{5}{25}-\frac{1}{25}\)

\(=\frac{4}{25}\)

b) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{101}{101}-\frac{1}{101}\)

\(=\frac{100}{101}\)

a) 1/5.6 + 1/6.7 + 1/7.8 + ... + 1/24.25

= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/24 - 1/25

= 1/5 - 1/25

= 4/25

b) 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 -1/101

= 1 - 1/101

= 100/101

c) 3/1.4 + 3/4.7 + ... + 3/2002.2005

= 1 - 1/4 + 1/4 - 1/7 + ... + 1/2002 - 1/2005

= 1 - 1/2005

= 2004/2005

d) 5/2.7 + 5/7.12 + ... + 5/1997.2002

= 1/2 - 1/7 + 1/7 - 1/12 + ... + 1/1997 - 1/2002

= 1/2 - 1/2002

= 500/1001

a,A =  \(\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{24\times25}\)

A\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

A\(=\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

b, B=\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)

B= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

B=\(1-\frac{1}{101}=\frac{100}{101}\)

c, \(C=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{2002\times2005}\)

C= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)

C= \(1-\frac{1}{2005}=\frac{2004}{2005}\)

d, D= \(\frac{5}{2\times7}+\frac{5}{7\times12}+...+\frac{5}{1997\times2002}\)

D= \(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{1997}-\frac{1}{2002}\)

D= \(\frac{1}{2}-\frac{1}{2002}=\frac{1001}{2002}-\frac{1}{2002}=\frac{1000}{2002}=\frac{500}{1001}\)

Câu 1:

a) \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)

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

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

b) \(B=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{4}{25}\)

Ta có : \(\dfrac{x^2}{5.6}\text{=}\dfrac{x^2}{5}-\dfrac{x^2}{6}\)

\(\dfrac{x^2}{6.7}\text{=}\dfrac{x^2}{6}-\dfrac{x^2}{7}\)

\(...\)

\(\dfrac{x^2}{24.25}\text{=}\dfrac{x^2}{24}-\dfrac{x^2}{25}\)

\(\Rightarrow\) biểu thức chỉ còn :

\(\dfrac{x^2}{5}-\dfrac{x^2}{25}\text{=}\dfrac{5x^2-x^2}{25}\text{=}\dfrac{4x^2}{25}\)