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( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750
( x + x + ... + x ) + ( 1 + 2 + ... + 100 ) = 5750
SSH là : ( 100 - 1) : 1 + 1 = 100 ( số )
Tổng là : ( 100 + 1 ) . 100 : 2 = 5050
=> 100x + 5050 = 5750
=> 100x = 700
=> x = 7
Vậy x = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : (x + 1) + (x + 2) + ..... + (x + 100) = 5750
=> (x + x + ...... + x) + (1 + 2 + ..... + 100) = 5750
=> 100x + 5050 = 5750
=> 100x = 700
=> x = 7 .
( x + 1 ) + ( x + 2 ) + ... + ( x + 100 ) = 5750
( x + x +.... + x ) + ( 1 + 2 + ... + 100 ) = 5750
x.100 + 5050 = 5750
x.100 = 5750 - 5050
x.100 = 700
x = 700 :100
x = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (2x-6)3 = (2x-6)2018
=> (2x-6)3 - (2x-6)2018 = 0
(2x-6)3.[1-(2x-6)2015 ] = 0
=> (2x-6)3 = 0 =>...
1 - (2x-6)2015 = 0 => (2x-6)2015 = 1 => ...
b) (2x-1)3 = 27 = 33
=> 2x - 1 = 3
=> ...
c) (x + 1) + (x+2) + (x+3) + ...+ (x+100) = 5750
x.100 + (1+2+3+...+100) = 5750
x.100 + [(1+100).100:2] = 5750
x.100 + 5050 = 5750
x.100 = 700
x = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
mình bik giải câu b thôi
TA CÓ
(x+1)+(x+2)+.................+(x+100)=5750
x+1+x+2+..............+x+100=5750
(x+x+....+x)+(1+2+3+.......+100)=5750
100.x + 5050 =5750
100.x=5750-5050
100x=700
=>x=700:100=7
VẬY X = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=1+2+22+23+…+220
=>2.A=2+22+23+24+…+221
=>2.A-A=2+22+23+24+…+221-1-2-22-23-…-220
=>A=221-1
Vậy 1+2+22+23+…+220=221-1
(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
Từ 1 đến 100 có:(100-1):1+1=100(số)
=>100.x+(100+1).100:2=5750
=>100.x+101.50=5750
=>100.x+5050=5750
=>100.x=5750-5050
=>100.x=700
=>x=7
Vậy x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
(x + x +.....+ x) +(1 + 2 +....+ 100)
100x + 5050=5750
100x=5750-5050=700
x=700:100=7
Vậy x = 7
(x+1)+(x+2)+(x+3)+.....+(x+100)=x+1+x+2+x+3+...+x+100=1+2+3+...100+100x=5050+100x=5750
100x=5750-5050=700
x=700/100=7
x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
x=7 nha Hà My
bài này là bài thi cuối HKI của bọn mk đấy
**** nha Hà My
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S=30+32+34+36+...+3200
6S=32+34+36+...+3202
6S-S=(32+34+36+...+3202)-(1+32+34+...+3200)
5S=1+(32-32)+(34-34)+...+(3200-3200)+3202
S=(3200+1):5\(\frac{ }{ }\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=100x+\left(1+2+3+4..+100\right)=5750\)
\(=100x+5050=5750\)
\(=100x=5750-5050=700\)
\(=x=700:100=7\)
\(x=7\)
(x + 1) + (x + 2) + (x + 3) + .... + (x + 100) = 5750
=> 100x + (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)
Bài giải
a, \(1075\cdot\left(x-3\right)\cdot\left(x-1\right)=0\)
\(\left(x-3\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{3\text{ ; }1\right\}\)
b, \(2\cdot\left(x-7\right)+3\cdot\left(x+1\right)\)
\(=2x-14+3x+3\)
\(=5x-11\)
c, \(x+1+x+2+...+x+100=5750\)
\(\left(x+x+...+x\right)+\left(1+2+...+100\right)=5750\)
\(100x+\left(100-1+1\right)\cdot\left(100+1\right)\text{ : }2=5750\)
\(100x+100\cdot101\text{ : }5=5750\)
\(100x+50\cdot101=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(x=700\text{ : }100\)
\(x=7\)