b) (x+1) + (x+2) + (x+3) +...+ (x+100)=5750
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.x\left(x+3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x+3=0\Leftrightarrow x=-3\end{cases}}\)
\(b,\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)
\(100x+\left(1+2+...+100\right)=5750\)
\(100x+\frac{100.101}{2}=5750\)
\(100x+5050=5750\)
\(100x=200\Leftrightarrow x=2\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(c,\)\(\left(x-1\right)+\left(x-2\right)+....+\left(x-100\right)=50\)
\(\left(x+x+...+x\right)-\left(1+2+...+100\right)=50\)
\(100x-5050=50\)
\(100x=50+5050\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)
\(a,\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+....+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(\Rightarrow x=7\)
\(b,x+\left(1+2+3+...+50\right)=2000\)
\(x+\frac{\left[1+50\right]\cdot\left[\left(50-1\right)\div1+1\right]}{2}=2000\)
\(x+1275=2000\)
\(\Rightarrow x=2000-1275=725\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)
\(\Leftrightarrow100x+5050=5750\)
\(\Leftrightarrow100x=700\)
hay x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
(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
![](https://rs.olm.vn/images/avt/0.png?1311)
(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
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)
\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)
\(100x-5050=50\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)
b. (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+4+...+100)=5750
=> 50 * x + 5050 = 5750
=> 50 * x = 5750-5050
=> 50 * x = 700
=> x = 700:50
=> x = 14
ây sorry.
từ dòng 4 bạn sửa lại :
=> 100 * x +5050 = 5750
=> ... tương tự
=> x=7