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21 tháng 4 2016

Ta có:

\(M=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(M=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(M=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(M=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(M=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}\right)\)

\(M=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}=N\)

\(\Rightarrow\frac{M}{N}=1\)

1:
- Ta có x^2+x+1=0
=> x^2+x=-1
=> x=x^2+1
mà x^2 x
=> x^2+1 x
=> Không tìm được giá trị của x
=> A không có giá trị

2.

Từ n2+n+1=0⇒n≠1⇒(n−1)(n2+a+1)=0⇒a3−1=0⇒a3=1n2+n+1=0⇒n≠1⇒(n−1)(n2+a+1)=0⇒a3−1=0⇒a3=1
Xét 3 trường hợp:
_ VỚi n = 3k thì A=(n3)k+1(n3)k=1+1=2(n3=1)A=(n3)k+1(n3)k=1+1=2(n3=1)
_ Với n = 3k + 1 thì A=(n3)k.n+1(n3)k.n=n+1n=n2+1n=−nn=−1A=(n3)k.n+1(n3)k.n=n+1n=n2+1n=−nn=−1
_Với n = 3k+2 thì A=(n3)k.n2+1(n3)k.n2=n2+1n2A=(n3)k.n2+1(n3)k.n2=n2+1n2
Ta có (n+1n)2=n2+1n2+2.n.1n=n2+1n2+2=1(n+1n)2=n2+1n2+2.n.1n=n2+1n2+2=1
 A = 1 -2 = -1
Mình không biết đúng không nha 

2 tháng 6 2016

ta có:

1/1.2+1/3.4+1/5.6+...+1/49.50

=>1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50

=>(1+1/3+1/5+1/7+...+1/49)-(1/2+1/4+1/6+...+1/50)

=>(1+1/2+1/3+...+1/49+1/50)-(1/2+1/4+1/6+...+1/50).2

=>(1+1/2+1/3+...+1/49+1/50) -( 1+1/2+1/3+...+1/25)

=>1/26+1/27+1/28+...+1/50=1/26+1/27+1/28+...+1/50

hay 1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50

2 tháng 6 2017

Ta có :

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\left(đpcm\right)\)

9 tháng 6 2017

Ta có

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

18 tháng 7 2015

1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50

=1/1-1/2+1/3-1/4+...+1/49-1/50

=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)

=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)

=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25

=1/26+1/27+...+1/50 (đpcm)

7 tháng 10 2016

bn ơi bn có thê

rhuowngs dẫn mình 

làm ko vì

mai mình ucngx

có bài này

27 tháng 7 2017

Câu hỏi của Phương Uyên - Toán lớp 7 | Học trực tuyến

mình ko có thời gian

bạn tự xem nhé

10 tháng 7 2016

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

=> đpcm

Ủng hộ mk nha ^_-

10 tháng 7 2016

đpcm là j z ạ

15 tháng 4 2016

A=1/1.2+1/3.4+.....+1/49.50

 =1-1/2+1/3-1/4+...+1/49-1/50=(1+1/3+1/5+...+1/49) - (1/2+1/4+1/6+...+1/50)

 =(1+1/3+1/5+...+1/49)+(1/2+1/4+1/6+...+1/50)-2.(1/2+1/4+1/6+...+1/50)

 =(1+1/2+1/3+1/4+...+1/49+1/50) - (1+1/2+1/3+...1/25)

 =1/26+1/27+...1/50

Vậy .........

3 tháng 7 2016

\(\frac{1}{1.2}+\frac{1}{3.4}+........+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+......+\frac{1}{50}-\left(1+\frac{1}{2}+....+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+........+\frac{1}{50}\)

\(\Rightarrowđpcm\)

3 tháng 7 2016

ta có:1/1.2+1/3.4+1/5.6+...+1/49.50=>1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50=>(1+1/3+1/5+1/7+...+1/49)-(1/2+1/4+1/6+...+1/50)=>(1+1/2+1/3+...+1/49+1/50)-(1/2+1/4+1/6+...+1/50).2=>(1+1/2+1/3+...+1/49+1/50) -( 1+1/2+1/3+...+1/25)=>1/26+1/27+1/28+...+1/50=1/26+1/27+1/28+...+1/50hay 1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50

Bấm mình nha

16 tháng 7 2017

Đặt \(A=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+....+\frac{1}{49\times50}\)

Dễ thấy\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{49}-\frac{1}{50}\)

Do đó

\(A=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{50}\right)\)

\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{49}+\frac{1}{50}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}\)

22 tháng 7 2017

cam on nha