(X + 1) + ( X + 3 ) +(X + 5 )+ …+ (X + 99) = 2550 
Số số hạng từ 1đến 99: (99-1):2+1= 50 ( số hạng) 
1+3+5+...+99= (99+1)x50: 2= 2500 
(X + 1) + ( X + 3 ) +(X + 5 )+ …+ (X + 99) = 50x X + 2500 =2550 
X = 2550 -2500 
X = 50

(x+1)+(x+3)+(x+5)+......+(x+99) = 2550

(x+x+x+....+x)+(1+3+5+....+99) = 2550

50x + 2500 = 2550

50x = 2550 - 2500

50x = 50

x = 50 : 50

x = 1

X=1 nhe

X = 1

chắc chắn 100%

<=> 50x + (1 + 3 + 5 + ....... + 99) = 2550

<=> 50x + 2500 = 2550

=> 50x = 2550 - 2500

=> 50x = 50

=> x = 50 : 50 '

=> x = 1

50x+(1+3+5+...+99)=2550

50x+2500               =2550

50x                        =2550-2500=50

    x                        =50:50=1

vậy x=1

( x + 1 ) + ( x + 3 ) + ( x + 5 ) + ... + ( x + 99 ) = 2550

( x + x + x + ... + x ) + ( 1 + 3 + 5 +... + 99 ) = 2550

 => 50x + ( 1 + 3 + 5 + ... + 99 ) = 2550

=> 50x + 2500 = 2550

=> 50x = 2550 - 2500

=> 50x = 50

x = 50 : 50

x = 1

Vậy x = 1

 (X + 1) + ( X + 3 ) +(X + 5 )+ …+ (X + 99) = 2550 
Số số hạng từ 1đến 99: (99-1):2+1= 50 ( số hạng) 
1+3+5+...+99= (99+1)x50: 2= 2500 
(X + 1) + ( X + 3 ) +(X + 5 )+ …+ (X + 99) = 50x X + 2500 =2550 
X = 2550 -2500 
X = 50

mk trả lời đầu tiên nhớ k nha!!

Tìm trung bình cộng của dãy số: 1;3;5;7;....;2009;2011

đáp số: X=50

\(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+.......\frac{1}{13x15}=\frac{1}{2}x\frac{2}{1x3}+\frac{2}{3x5}.......+\frac{2}{13x15}\)

\(A=\frac{1}{2}x\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\right)\)

Còn lại em nhân giống ở trên nhé

Đặt A = 1/15 + 1/35 + ... + 1/3135 

       A = 1/3.5 + 1/5.7 + ... + 1/55.57

     2A =  2/3.5 + 2/5.7 + ... + 2/55.57 

    2A = 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/55 - 1/57 

    2A = 1/3 - 1/57 = 6/19 

      A = 3/19 

thanks các bn

a: \(x+6\dfrac{1}{8}=8\)

=>\(x+\dfrac{49}{8}=\dfrac{64}{8}\)

=>\(x=\dfrac{64}{8}-\dfrac{49}{8}=\dfrac{15}{8}\)

b: \(\dfrac{11}{2}\cdot x=\dfrac{1}{5}:\dfrac{1}{3}\)

=>\(x\cdot\dfrac{11}{2}=\dfrac{1}{5}\cdot3=\dfrac{3}{5}\)

=>\(x=\dfrac{3}{5}:\dfrac{11}{2}=\dfrac{3}{5}\cdot\dfrac{2}{11}=\dfrac{6}{55}\)

c: \(x\cdot\dfrac{3}{5}+\dfrac{2}{5}\cdot x=\dfrac{4}{9}+\dfrac{1}{3}\)

=>\(x\left(\dfrac{3}{5}+\dfrac{2}{5}\right)=\dfrac{4}{9}+\dfrac{3}{9}\)

=>\(x\cdot1=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}\)

a) \(\frac{1}{3}+\frac{5}{6}:\left(x-2\frac{1}{5}\right)=\frac{3}{4}\)

=> \(\frac{1}{3}+\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}\)

=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}-\frac{1}{3}\)

=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{5}{12}\)

=> \(x-\frac{11}{5}=\frac{5}{6}:\frac{5}{12}\)

=> \(x-\frac{11}{5}=2\)

=> \(x=2+\frac{11}{5}\)

=> \(x=\frac{21}{5}\)

thanks bn

1/15 +1/35 + 1/63 + ... + 1/3135

= 1/3*5 + 1/5*7 + 1/7*9 + ... + 1/55*57

= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/55 - 1/57

= 1/3 - 1/57

= 18/57

1/3 + 1/6 + 1/10 +  ... + 1/x(x + 1)*2 = 2009/2011

=> 2/6 + 2/12 + 2/20 + ... + 2/x(x + 1)*2 = 2009/2011

=> 2(1/6 + 1/12 + 1/20 + ... + 1/x(x + 1)) = 2009/2011

=> 2(1/2*3 + 1/3*4 + 1/4*5 + ... + 1/x(x + 1)) = 2009/2011

=> 2(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x + 1)  = 2009/2011

=> 2(1/2 - 1/x + 1) = 2009/2011

=> 1/2 - 1/x + 1 = 2009/4022

=> 1/x + 1 = 1/2011

=> x + 1 = 2011

=> x = 2010

vậy-

Phanh đã chuyển đến đây 

1/2 . x + 3/5 . (x-2) = 3

1/2 . x + 3/5 . x - 3/5 . 2 = 3

1/2 . x + 3/5 . x - 6/5 = 3

x.(1/2+3/5) - 6/5 = 3

x.11/10 - 6/5 = 3

x.11/10 =3+6/5

x.11/10 =21/5

x = 21/5:11/10

x =42/11 

này làm gì thế,nghĩa là sao

`x xx1/2+x/2+x xx 99=0,25`

`x xx 1/2+x :2+x xx 99=0,25`

`x xx 1/2+x xx1/2+x xx 99=0,25`

`x xx (1/2+1/2+99)=0,25`

`x xx 100=0,25`

`x=0,25:100`

`x=0,0025`