\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

\(=1-\dfrac{1}{25}\)

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

Nhanh giúp mình với cả nhà ơi

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2006.2007}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2006}-\frac{1}{2007}\)

\(=1-\frac{1}{2007}\)

\(=\frac{2006}{2007}\)

CẦN GẤP NHÉ MỌI NGƯỜI

)chứng tỏ

a)1/1x2+1/2x3+...+1/9x10 <1

b)1/1x2+1/2x3+...+1/99x100 <1

a)4/1x5+1/5x9+1/9x13+1/13x17+1/17x21<1

Lưu ý:"x" là phép nhân

Toán lớp 6

ái tích mình tíc lại nhà 

CÂU a đề bài nó sao sao đó

mà gợi ý cho bạn ....bạn tính tổng đó ra bao nhiêu rồi đem so sánh cho 1

1/1 x 2 + 1/2 x 3 + 1/3 x 4 + .... + 1/9 x 10

= 1 - 1/2 + 1/2 - 1/3 +1/3 - 1/4 + ... + 1/9 - 1/10

= 1 - 1/10

= 9/10

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{9\times10}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\\ =1-\dfrac{1}{10}\\ =\dfrac{9}{10}\)

9/10 nha

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100=99/100

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100

= 99/100

\(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\) + ...+ \(\dfrac{1}{x\times\left(x+1\right)}\)  = \(\dfrac{1}{2}\)

\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) +...+ \(\dfrac{1}{x}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{1}{2}\)

1 - \(\dfrac{1}{x+1}\) = \(\dfrac{1}{2}\)

       \(\dfrac{1}{x+1}\) = 1 - \(\dfrac{1}{2}\)

       \(\dfrac{1}{x+1}\) = \(\dfrac{1}{2}\)

       \(x+1\)  = 1 : \(\dfrac{1}{2}\)

        \(x\) + 1  = 2

        \(x\)         = 2 - 1

         \(x\)        = 1 

nhanh lên mọi người ơi , mình đang vội☹

1/1x2 + 1/2x3 + 1/3x4 +1/4x5 +1/5x6

= 1 -1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6

= 1 - 1/6 = 5/6

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{100\cdot101}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{100}-\dfrac{1}{101}\)

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

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