\(\left[x+1\right]+\left[x+2\right]+\left[x+3\right]+...+\left[x+100\right]=5050\)

\(\Rightarrow x+1+x+2+x+3+...+x+100=5050\)

\(\Rightarrow(x+x+...+x)+(1+2+3+...+100)=5050\)

\(\Rightarrow100x+5050=5050\)

\(\Rightarrow100x=5050-5050\)

\(\Rightarrow100x=0\)

\(\Rightarrow x=\frac{0}{100}=0\)

\(\Leftrightarrow100x+5050=5050\Rightarrow x=0\)

(x+1)+(x+2)+(x+3)+...........+(x+100)=5050

x+1+x+2+x+3+.............+x+100=5050

x*100+1+2+3+.............+100=5050

x*100+5050=5050

x*100=5050-5050=0

x=0/100=0

(x+1)+(x+2)+(x+3)+........+(x+100)=5050

=> x+1+x+2+x+3+...+x+100=5050

=> (x+x+...+x)+(1+2+3+...+100)=5050

=> 100*x+5050=5050

=> 100*x=5050-5050

=> 100*x=0

=> x=0:100

=> x=0

x = 0 nha bn

Chúc bn học tốt !!!

1 + 2 + 3 + ... + n = 5050

\(\Rightarrow\frac{\left(n+1\right)}{2}\) .(n−1+1)=5050 (trung bình cộng nhân cho số các số hạng)

\(\Rightarrow\frac{\left(n+1\right)}{2}\) .n=5050

\(\Rightarrow\frac{n\left(n+1\right)}{2}\) =5050

\(\Rightarrow\)n(n+1)=5050.2

\(\Rightarrow\)n(n+1)=10100

\(\Rightarrow\)n(n+1)=100.101

Vậy n=100

1 + 2 + 3 + ... + n = 5050

=> ( 1 + n ) . n ÷ 2 = 5050

=> ( 1 + n ) . n = 10100

mà 1 + n và n là 2 số tự nhiên liên tiếp.

mà 10100 = 100 .101

vậy n bằng 100

Tìm n:

1 + 2 + 3 + ... + n = 5050

\(\frac{\left(n+1\right)}{2}.\left(n-1+1\right)=5050\)

\(\frac{\left(n+1\right)}{2}.n=5050\)

\(\frac{n\left(n+1\right)}{2}=5050\)

\(n\left(n+1\right)=5050.2\)

\(n\left(n+1\right)=10100\)

\(n\left(n+1\right)=100.101\)

Vậy \(n=100\)

(X + 1) + (X + 2) + (X + 3) + … + (X + 99) + (X +100) = 5050

(X+X+X+...+X+X) + (1+100) x 50 = 5050 

100 x X + 101 x 50 = 5050

100 x X + 5050 = 5050

100 x X = 5050 - 5050

100 x X = 0

X= 0: 100

X= 0

`(x + 1) + (x + 2) + (x + 3) + … + (x + 99) + (x +100) = 5050`

`(x+x+x+...+x+x) + (1+2+3+...+99+100)=5050`

Số số hạng là : `(100-1)/1+1= 100` ( số hạng )

Tổng dãy là : `(100+1) xx 100 : 2=5050`

`100x+5050=5050`

` 100x=5050-5050`

` 100x=0`

`x=0`

100x+(1+2+3+...+100)=5050

100x+5050=5050

100x=5050-5050

100x=0

x=0:100

x=0

a, (x+1)+(x+2)+(x+3)+...+(x+100) = 7450

(x+x+...+x)+(1+2+...+100) = 7450

100 x + 101 . 100 2 = 7450

100x = 2400

x = 24

b, 1+2+3+...+x = 500500

Đặt: A = 1+2+3+...+x

số hạng A (x - 1) : 1 + 1 = x

Tổng của A

A = x + 1 . x 2 = 500500

(x+1).x = 1001000

Ta thấy

1000.1001 = 1001000

=> x = 1000

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\cdot\cdot\cdot+\left(x+100\right)=5750\)

\(\Rightarrow\left(x+x+x+\cdot\cdot\cdot+x\right)+\left(1+2+3+\cdot\cdot\cdot+100\right)=5750\) (1)

Đặt \(A=1+2+3+\cdot\cdot\cdot+100\)

Số các số hạng trong \(A\) là: \(\left(100-1\right):1+1=100\left(số\right)\)

Tổng \(A\) bằng: \(\left(100+1\right)\cdot100:2=5050\)

Khi đó, (1) trở thành:

\(100x+5050=5750\)

\(\Rightarrow100x=5750-5050\)

\(\Rightarrow100x=700\)

\(\Rightarrow x=\dfrac{700}{100}=7\left(tm\right)\)

Vậy \(x=7\).

(x+1)+(x+2)+...+(x+100)=5750

=>x+x+...+x+1+2+3+...+100=5750

=>100x+5050=5750

=>100x=700

=>x=7

a)<=> 3x-5-x=0

   <=>    2x-5=0

   <=>        x=5/2

c) x.(1+2+3+4+...+100)=0

    x.5050=0

     x=0:5050=0

Vậy x=0

d) x.(1+2+3+4+5+...+100)=5050

    x.5050=5050

    x=1

Vậy x=1

e) x+1+x+2+x+3+x+4+...+x+100=5050

    (x+x+x+x+...+x)+(1+2+3+4+...+100)=5050

     100 số hạng x

    x.100+5050=5050

    x.100=0

    x=0

Vậy x=0

a , x = \(\frac{5}{2}\)

=>100x+1+2+..+100=5750

=>100x+\(\frac{\left(100+1\right)100}{2}\)=5750

=>100x+5050=5750

=>100x=700

=>x=7