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18 tháng 11 2018

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 49 − 1 2 + 1 4 + 1 6 + ⋯ ⋯ + 1 50 = 1 + 1 2 + 1 3 + 1 4 + 1 5 + ⋯ ⋯ + 1 50 − 2. 1 2 + 1 4 + 1 6 + ⋯ ⋯ + 1 50 = 1 + 1 2 + 1 3 + 1 4 + 1 5 + ⋯ ⋯ + 1 50 − 1 + 1 2 + 1 3 + ⋯ ⋯ + 1 25 = 1 26 + 1 27 + 1 28 + ... + 1 50       ( d p c m )

25 tháng 7 2016

Ta có : 

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

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

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

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

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

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

=> đpcm

Ủng hộ mk nha !!! ^_^

30 tháng 7 2016

\(\text{Ta có :}\)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

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

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

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

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

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

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

12 tháng 10 2018

Ta có :

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

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

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

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)

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

    Vậy ......

                         ~Hok tốt~

12 tháng 10 2018

CỨ TÍNH VẾ TRÁI R ĐC 

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

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

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

(1+1/2+1/4+1/6+...+1/50)-(1+1/2+1/3+..+1/25)

1/26+1/27+...+1/50

15 tháng 1 2017

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

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

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

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

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

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+...+\frac{1}{50}\left(đpcm\right)\)

15 tháng 1 2017

cảm ơn bn nha ^.^!!!

27 tháng 3 2019

\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}\)

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

\(=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+.....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+.....+\frac{1}{50}\right)\)

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

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

27 tháng 3 2019

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

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

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

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

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

30 tháng 11 2015

Mới nhìn mà muốn vào viện ngay rồi