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Bài 10:
a: 2x-3 là bội của x+1
=>\(2x-3⋮x+1\)
=>\(2x+2-5⋮x+1\)
=>\(-5⋮x+1\)
=>\(x+1\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{0;-2;4;-6\right\}\)
b: x-2 là ước của 3x-2
=>\(3x-2⋮x-2\)
=>\(3x-6+4⋮x-2\)
=>\(4⋮x-2\)
=>\(x-2\inƯ\left(4\right)\)
=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{3;1;4;0;6;-2\right\}\)
Bài 14:
a: \(4n-5⋮2n-1\)
=>\(4n-2-3⋮2n-1\)
=>\(-3⋮2n-1\)
=>\(2n-1\inƯ\left(-3\right)\)
=>\(2n-1\in\left\{1;-1;3;-3\right\}\)
=>\(2n\in\left\{2;0;4;-2\right\}\)
=>\(n\in\left\{1;0;2;-1\right\}\)
mà n>=0
nên \(n\in\left\{1;0;2\right\}\)
b: \(n^2+3n+1⋮n+1\)
=>\(n^2+n+2n+2-1⋮n+1\)
=>\(n\left(n+1\right)+2\left(n+1\right)-1⋮n+1\)
=>\(-1⋮n+1\)
=>\(n+1\in\left\{1;-1\right\}\)
=>\(n\in\left\{0;-2\right\}\)
mà n là số tự nhiên
nên n=0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\cdot\cdot\cdot+\left(x+100\right)=5750\)
\(\Rightarrow\left(x+x+x+\cdot\cdot\cdot+x\right)+\left(1+2+3+\cdot\cdot\cdot+100\right)=5750\) (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
![](https://rs.olm.vn/images/avt/0.png?1311)
=>100x+1+2+..+100=5750
=>100x+\(\frac{\left(100+1\right)100}{2}\)=5750
=>100x+5050=5750
=>100x=700
=>x=7
![](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: =>41-(2x-5)=720:40=18
=>2x-5=23
=>2x=28
=>x=14
b: =>100x+5050=5750
=>100x=700
=>x=7
a: =>41-(2x-5)=720:40=18
=>2x-5=23
=>2x=28
=>x=14
b: =>100x+5050=5750
=>100x=700
=>x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Rightarrow\left(x+x+x+...+x\right)+\left(1+2+3+..+100\right)=5750\Rightarrow x.100+\left(100+1\right)\cdot100:2=5750\)\
\(\Rightarrow x.100+5050=5750\Rightarrow x.100=700\Rightarrow x=7\)
b) \(\frac{x+1}{2}=\frac{8}{x+1}\Rightarrow\left(x+1\right)\left(x+1\right)=2.8\)
\(\Rightarrow\left(x+1\right)^2=16\Rightarrow\left(x+1\right)^2=4^2\)
\(\Leftrightarrow x+1=4\Rightarrow x=3\)
1.\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(\Leftrightarrow100x+5050=5750\)
\(\Leftrightarrow100x=5750-5050=700\)
\(\Leftrightarrow x=700:100=7\)
2. \(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=8.2\)
\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=16\)
\(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow\left(x+1\right)=16:2\)
\(\Leftrightarrow\left(x+1\right)=8\)
\(\Leftrightarrow x=8-1=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(100-7\left(x-5\right)=58\)
\(\Leftrightarrow7\left(x-5\right)=42\)
\(\Leftrightarrow x-5=6\)
hay x=11
b: Ta có: \(12\left(x-1\right):3=4^3+2^3\)
\(\Leftrightarrow12\left(x-1\right)=216\)
\(\Leftrightarrow x-1=18\)
hay x=19