1/1.2 + 1/2.3 + ...... + 1/49.50

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

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

\(-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)

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 !!! ^_^

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

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

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

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

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

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

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

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

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

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

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

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

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 

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

Ta biến đổi vế phải :

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

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

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

Vậy 1/26+1/27+1/28+.....+1/50=1-1/2+1/3-1/4+......+1/49-1/50

Mình không bấm phân số được mong mấy bạn thông cảm

\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)

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

\(A=\frac{1}{1}-\frac{1}{50}\)

\(A=\frac{49}{50}\)

Vì \(\frac{245}{420}< \frac{245}{294}< \frac{245}{250}\)

Vậy \(\frac{7}{12}< \frac{49}{50}< \frac{5}{6}\)