(5/5x10+5/10x15+5/15x20+...+5/95x100)

=  (1/5-1/10+1/10-1/15+1/15-1/20+....+1/95-1/100):5

=(1/5-1/100):5

=19/500

Đặt A = 1/5x10 + 1/10x15 + 1/15x20 + 1/20x25 + ... + 1/95x100

A x 5 = 5/5x10 + 5/10x15 + 5/15x20 + 5/20x25 + ... + 5/95x100

A x 5 = 1/5 - 1/10 + 1/10 - 1/15 + 1/15 - 1/20 + 1/20 - 1/25 + ... + 1/95 - 1/100

A x 5 = 1/5 - 1/100

A x 5 = 19/100

A = 19/100 : 5

A = 19/100 x 1/5 

A = 19/500

 Vậy A= 19/500

chỉ cần phân k nha bạn 

kết bạn vs mình nói cho

các bạn tíck cho mình thì sẽ trả cho 3 tíck

N = 1/1x5 + 1/5x10 + 1/10x15 + 1/15x20 + .....+1/2005 x 2010

N = 1 - 1/5 +1/5-1/5+1/10-1/15+1/5-1/20+.....+1/2005-1/2010

N = 1 - 1/2010

N = 2009/2010

Ta có:

\(N=\frac{1}{1x5}+\frac{1}{5x10}+\frac{1}{10x15}...+\frac{1}{2005x2010}\)

\(\Rightarrow Nx5=\left(\frac{1}{1x5}+\frac{1}{5x10}+\frac{1}{10x15}...+\frac{1}{2005x2010}\right)x5\)

\(=\frac{5}{1x5}+\frac{5}{5x10}+\frac{5}{10x15}...+\frac{5}{2005x2010}\)

\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2005}-\frac{1}{2010}\)

\(=1-\frac{1}{2010}\)

\(=\frac{2009}{2010}\)

\(\Rightarrow N=\frac{2009}{2010}:5=\frac{2009}{2010}x\frac{1}{5}=\frac{2009}{10050}\)

Bạn thuỳ dung ơi cho hỏi sao 5/(1x5) = 1 - 1/5 vậy ???

Ta có:

\(\frac{5}{1\cdot7}+\frac{5}{7\cdot13}+\frac{5}{13\cdot19}+...+\frac{5}{91\cdot97}\)

= \(5\cdot\frac{1}{6}\cdot\left(\frac{6}{1\cdot7}+\frac{6}{7\cdot13}+\frac{6}{13\cdot19}+...+\frac{6}{91\cdot97}\right)\)

= \(\frac{5}{6}\cdot\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+...+\frac{1}{91}-\frac{1}{97}\right)\)

= \(\frac{5}{6}\cdot\left(1-\frac{1}{97}\right)\)

= \(\frac{5}{6}\cdot\frac{96}{97}\)

= \(\frac{80}{97}\)

5/1.7 + 5/7.13 + 5/13.19 + ... + 5/91.97

= 5/6.(1 - 1/7 + 1/7 - 1/13 + 1/13 - 1/19 + ... + 1/91 - 1/97)

= 5/6.(1 - 1/97)

= 5/6.96/97

= 80/97

5/1.7+5/7.13+5/13.19+....+5/91.97

=5.(1/1.7+1/7.13+1/13.19+...+1/91.97)

=5.1/6.(6/1.7+6/7.13+6/13.19+....+6/91.97)

=5/6.(1-1/7+1/7-1/13+1/13-1/19+.....+1/91-1/97)

=5/6.(1-1/97)

=5/6.96/97

=80/97

Hok tốt

(x-5)⁴ = (x-5)⁶

(x-5)⁴ - ( x-5)⁶ = 0

(x-5)⁴.{ 1 - (x-5)² }= 0

\(\left[{}\begin{matrix}x-5=0\\(x-5)^2=1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\x-5=1\\x-5=-1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

\(x\in\) { 4; 5; 6}

\(=5\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\right)\)

\(=5\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)\)

\(=5\cdot\dfrac{5}{24}=\dfrac{25}{24}\)

`5/12 + 5/20 + 5/30 + 5/42 + 5/56`

`= 5/(3.4) + 5/(4.5) + 5/(5.6) + 5/(6.7) + 5/(7.8)`

`= 5 . (1/(3.4) + 1/(4.5) + ... + 5/(7.8))`

`= 5 . (1/3 - 1/4 + 1/4 - 1/5 + ...+ 1/7 - 1/8)`

`= 5 . (1/3 - 1/8) `

`= 5 . 5/24 = 25/24`

`S_1 = 5/(1.4) + 5/(4.7) +...+ 5/(97.100)`

`S_1 = 5 (1/(1.4) + 1/(4.7) +...+ 1/(97.100))`

`S_1 = 5/3 (3/(1.4) + 3/(4.7) +...+ 3/(97.100))`

`S_1 = 5/3 (1 - 1/4 + 1/4 - 1/7 + ...+ 1/97 - 1/100)`

`S_1 = 5/3 (1 - 1/100)`

`S_1 = 5/3 . 99/100`

`S_1 = 33/20`