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12 tháng 3 2015

mình trả lời bài 1 thôi nhé :

Gọi biểu thức trên là A.

Theo bài ra ta có:A=1/1.6+1/6.11+1/11.16+...+1/(5n+1)+1/(5n+6)

                           =1/5(1-1/6+1/6-1/11+1/11-1/16+...+1/5n+1-1/5n+6)

                           =1/5(1-1/5n+6)

                           =1/5( 5n+6/5n+6-1/5n+6)

                           =1/5(5n+6-1/5n+6)

                           =1/5.5n+5/5n+6

                           =n+1/5n+6

                           =ĐIỀU PHẢI CHỨNG MINH

 

30 tháng 4 2015

x- 20/11.13 - 20/13.15 - 20/13.15 - 20/15.17 -...- 20/53.55=3/11

x-10.(2/11.13+2/13.15+2/15.17+...+2/53.55=3/11

x-10.(1/11-1/13+1/13-1/15+1/15-1/17+...+1/53-1/55)=3/11

x-10.(1/11-1/55)=3/11

x-10.4/55=3/11

x-8/11=3/11

x = 3/11+8/11

x=11/11=1

****

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

=1/2-1/100=49/100

b; =5/3(1-1/4+1/4-1/7+...+1/100-1/103)

=5/3*102/103

=510/309=170/103

c: =1/2(1/3-1/5+1/5-1/7+...+1/49-1/51)

=1/2*16/51=8/51

8 tháng 8 2023

 CM:  \(\dfrac{1}{1.6}\)\(\dfrac{1}{11.16}\)+...+ \(\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\) = \(\dfrac{n+1}{5n+6}\)

A = \(\dfrac{1}{5}\)(\(\dfrac{5}{1.6}\) + \(\dfrac{5}{6.11}\)+...+ \(\dfrac{5}{\left(5n+1\right).\left(5n+6\right)}\)

A = \(\dfrac{1}{5}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{6}\)\(\dfrac{1}{6}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{5n+1}\) - \(\dfrac{1}{5n+6}\))

A = \(\dfrac{1}{5}\) .( \(\dfrac{1}{1}\) - \(\dfrac{1}{5n+6}\))

A = \(\dfrac{1}{5}\)\(\dfrac{5n+6-1}{5n+6}\)

A = \(\dfrac{1}{5}\)\(\dfrac{5n+5}{5n+6}\)

A = \(\dfrac{1}{5}\) . \(\dfrac{5.\left(n+1\right)}{5n+6}\)

A = \(\dfrac{n+1}{5n+6}\)

\(\dfrac{1}{1.6}\) + \(\dfrac{1}{6.11}\)\(\dfrac{1}{11.16}\)+...+ \(\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\) = \(\dfrac{n+1}{5n+1}\) (đpcm)

 

 

8 tháng 8 2023

\(A=\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\)

\(A=\dfrac{1}{5}\left[1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right]\)

\(A=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)

\(A=\dfrac{1}{5}\left(\dfrac{5n+6-1}{5n+6}\right)=\dfrac{1}{5}\left(\dfrac{5n+5}{5n+6}\right)=\dfrac{1}{5}.5\left(\dfrac{n+1}{5n+6}\right)=\dfrac{n+1}{5n+6}\)

\(\Rightarrow dpcm\)

\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)

\(=\dfrac{1}{5}\cdot\dfrac{5n+6-1}{5n+6}\)

\(=\dfrac{n+1}{5n+6}=VP\)

14 tháng 2 2020

Không ai giúp tôi giải mấy bài này à?

28 tháng 2 2022

EM LỚP 5 KO BT LÀM

HÌ (>_<)