\(a.x\left(x+3\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x+3=0\Leftrightarrow x=-3\end{cases}}\)

\(b,\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)

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

\(100x+\frac{100.101}{2}=5750\)

\(100x+5050=5750\)

\(100x=200\Leftrightarrow x=2\)

a) \(x.\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

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

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

                                   100x + 5050 = 5750

                                                 100x = 5750 - 5050

                                                  100x = 700

                                                        x = 700 : 100 = 7

a,\(X.\left(x+3\right)=0\)

điều kiện xảy ra 2 trường hợp

trường hợp 1:\(x=0\)

trường hợp 2:\(x+3=0\)

\(\Rightarrow x=0-3=-3\)

b,từ 1 đến 100 có 100 số x

vậy ta loại 100x ra một vế được:

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

tổng:\(100.101:2=5050\)

\(\Rightarrow100x=200\)

\(\Rightarrow x=2\)

\(c,\)\(\left(x-1\right)+\left(x-2\right)+....+\left(x-100\right)=50\)

\(\left(x+x+...+x\right)-\left(1+2+...+100\right)=50\)

\(100x-5050=50\)

\(100x=50+5050\)

\(100x=5100\)

\(\Rightarrow x=\frac{5100}{100}=51\)

\(a,\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+....+\left(x+100\right)=5750\)

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

\(100x+5050=5750\)

\(100x=5750-5050\)

\(100x=700\)

\(\Rightarrow x=7\)

\(b,x+\left(1+2+3+...+50\right)=2000\)

 \(x+\frac{\left[1+50\right]\cdot\left[\left(50-1\right)\div1+1\right]}{2}=2000\)

\(x+1275=2000\)

\(\Rightarrow x=2000-1275=725\)

c , ( x - 1 ) + ( x - 2 ) + ...... + ( x - 100 ) = 50

( x + x + ........ + ) - ( 1 + 2 + ..... + 100 ) = 50

100x - 5050 = 50

100x = 50 + 5050

100x = 5100

=> x = 5100/100 = 51

Ta có: \(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)

\(\Leftrightarrow100x+5050=5750\)

\(\Leftrightarrow100x=700\)

hay x=7

x=7

tham kharo : https://olm.vn/hoi-dap/detail/89184552036.html

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

(x+x+...+x)+(1+2+3+...+100)=5750

100x+5050=5750

100x=5750-5050

100x=700

x=700/100

x=7

\(\Leftrightarrow100x+5050=5750\)

hay x=7

(x+1) + (x+2) + (x+3) +....+ (x+100) = 5750

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

(x + x + x +...+ x) + (1 + 2 + 3 +...+ 100) = 5750

100x + 5050 = 5750

100x = 5750 - 5050

100x = 700

x = 700 : 100

x = 7

=>100x+5050=5750

=>100x=700

=>x=7

a) Dãy trên có số số hạng là :

( x - 1 ) : 1 + 1 = ( x - 1 ) + 1 = x 

=> 1 + 2 + 3 + 4 + ... + x = 45

=> ( x + 1 ) . x : 2  = 45

=> ( x + 1 ) . x = 90

=> ( x + 1 ) . x = 10 . 9

=> x = 9

Vậy x = 9

a) Dãy trên có số số hạng là :

( x - 1 ) : 1 + 1 = ( x - 1 ) + 1 = x 

=> 1 + 2 + 3 + 4 + ... + x = 45

=> ( x + 1 ) . x : 2  = 45

=> ( x + 1 ) . x = 90

=> ( x + 1 ) . x = 10 . 9

=> x = 9

Vậy x = 9

b) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750

=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750 ( mỗi nhóm có 100 số hạng )

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

=> 100x + 5050 = 5750

=> 100x = 700

=> x = 7

Vậy x = 7

A.   \(\left(x+1\right)+\left(x+2\right)+......+\left(x+100\right)=5750\)

      \(x+1+x+2+....+x+100=5750\)

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

      \(100x+5050=5750\)

\(100x=700\)

\(x=700:100=7\)

B.   x+(1+2+......+100) = 2000

       x + 5050 = 2000

            x = 2000 - 5050

           x= -3050

C.   ( x-1 )+(x-2)+......+( x - 100 ) = 50

 x-1+x-2+.........+x-100 = 50

100x + ( -1-2-........-100  ) = 50

100x + ( - 5050 ) = 50

100x = 50 + 5050

100 x = 5100

x = 5100 : 100

x = 51

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

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

\(100x+5050=5750\)

\(100x=5750-5050\)

\(100x=700\)

\(\Rightarrow x=\frac{700}{100}=7\)

B. \(x+\left(1+2+3+4+5+....+100\right)=2000\)

 \(x+\frac{\left(100+1\right).100}{2}=2000\)

\(x+5050=2000\)

\(\Rightarrow x=2000-5050=-3050\)

C. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)

\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)

\(100x-5050=50\)

\(100x=5100\)

\(\Rightarrow x=\frac{5100}{100}=51\)