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1) Do x ∈ Z và 0 < x < 3
⇒ x ∈ {1; 2}
2) Do x ∈ Z và 0 < x ≤ 3
⇒ x ∈ {1; 2; 3}
3) Do x ∈ Z và -1 < x ≤ 4
⇒ x ∈ {0; 1; 2; 3; 4}
![](https://rs.olm.vn/images/avt/0.png?1311)
1) (x-1)(x+5)(-3x+8)=0
\(\hept{\begin{cases}\\\\\end{cases}}\)
1) (x-1)(x+5)(-3+8)=0
= (x-1)(x+5).5 =0
\(\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0+1=1\\x=0-5=-5\end{cases}}\)
\(\Rightarrow x\in\left\{1;-5\right\}\)
2) (x-1)(x-2)(x-3)=0
\(\hept{\begin{cases}x-1=0\\x-2=0\\x-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0+1=1\\x=0+2=2\\x=0+3=3\end{cases}}\)
\(\Rightarrow x\in\left\{1;2;3\right\}\)
3)(5x+3)(x2+4)(x-1)=0
\(\hept{\begin{cases}5x+3=0\\x^2+4=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}5x=0-3=-3\\x^2=0-4=-4\\x=0+1=1\end{cases}}\Rightarrow\hept{\begin{cases}x=-3:5\Rightarrow x\in\varnothing\\x\in\varnothing\\x=1\end{cases}}\)
\(\Rightarrow x=1\)
4)x(x2-1)=0
\(\orbr{\begin{cases}x=0\\x^2-1=0\Rightarrow x^2=0+1=1\Rightarrow x^2=1^2;(-1)^2\Rightarrow x\in\left\{1;-1\right\}\end{cases}}\)
\(\Rightarrow x\in\left\{-1;0;1\right\}\)
Xin lỗi về phần bên trên nha! tại tui ấn nhầm nút.Sorry.
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(2x=0\)
\(\Leftrightarrow x=0\)
Vậy ....
b/ \(\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy ...
c/ \(\left(x-2\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)
Vậy ....
d/ \(\left(x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-1\right)^2=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
b) \(x\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
c) \(\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
d) \(\left(x+5\right)\left(x^2+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\x^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x^2=-1\end{cases}}\Rightarrow x=-5\)
a) ( x - 1 )2 = 0
x - 1 = 0
x = 1
b) x . ( x - 1 ) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
c) ( x + 1 ) . ( x - 2 ) = 0
tương tự, xét 2 trường hợp như câu b
d) tương tự, xét 2 trường hợp như câu b
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\left(3-x\right)\left|x+5\right|=0\)
\(\Leftrightarrow\orbr{\begin{cases}3-x=0\\\left|x+5\right|=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
Ta có :
\(\left(3-x\right)\left|x+5\right|=0\)
\(\Rightarrow\orbr{\begin{cases}3-x=0\\\left|x+5\right|=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x+5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x\left(x+2\right)=0\)
=> \(\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
Vậy ...
b) (x - 1)(x - 2) = 0
=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
Vậy ...
còn lại tương tự
a) x(x+2) = 0
=> x=0 hoặc x+2 = 0
+ x + 2 = 0
x = 0 - 2
x = -2
Vậy x thuộc tập hợp 0 ; -2
b) (x-1)(x-2)=0
=> x-1 =0 hoặc x-2=0
+ x-1=0 + x-2=0
x=0+1 x=0+2
x=1 x=2
Vậy x thuộc tập hợp 1;2
ý c , d làm giống 2 ý đầu
e) x(x-3)>0
=> \(\hept{\begin{cases}x>0\\x-3>0\end{cases}}\)=> \(\hept{\begin{cases}x>0\\x>3\end{cases}}=>\hept{\begin{cases}\\\end{cases}}x>3\)
Vậy x > 3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Rightarrow x=1\)
\(x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
\(\left(x-1\right)^2=0\)
\(\left(x-1\right)^2=0^2\)
\(\Rightarrow x-1=0\)
\(x=0-1\)
\(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:tìm x thuộc Z
a)x.(x-1)=0
\(\Leftrightarrow\left[\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy: \(x=0;1\)
b)(x-3).(x+4)=0
\(\Leftrightarrow\left[\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
Vậy: \(x=3;-4\)
c)(2x-4).(x+2)=0
\(\Leftrightarrow2\left(x-2\right).\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x=2;-2\)
d)(x+1)^2.(x-2)^2=0
\(\Leftrightarrow\left[\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy: \(x=-1;2\)
e) x(x+1).(x+2)^2.(x+3)^3=0
\(\Leftrightarrow\left[\begin{matrix}x=0\\x+1=0\\x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=-1\\x=-2\\x=-3\end{matrix}\right.\)
Vậy: \(x=0;-1;-2;-3\)
f)(x-9)^5.(x-5)^8=0
\(\Leftrightarrow\left[\begin{matrix}x-9=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=9\\x=5\end{matrix}\right.\)
Vậy: \(x=9;5\)
g)x(x+100)^10.(x+2000)^20.(x+300)^300=0
\(\Leftrightarrow\left[\begin{matrix}x=0\\x+100=0\\x+200=0\\x+300=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0\\x=-100\\x=-200\\x=-300\end{matrix}\right.\)
Vậy: \(x=0;-100;-200;-300\)
h)(x-2)^2=0
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy: \(x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: (x-1)(x-2)<=0
=>1<=x<=2
mà x là số nguyên
nên \(x\in\left\{1;2\right\}\)
2: \(\left(2x-4\right)\left(2x-10\right)< 0\)
=>4<2x<10
=>2<x<5
mà x là số nguyên
nên \(x\in\left\{3;4\right\}\)
4: \(\left(x^2-7\right)\left(x^2-1\right)< =0\)
\(\Leftrightarrow1\le x^2\le7\)
mà x là số nguyên
nên \(x\in\left\{1;-1;2;-2\right\}\)
Tìm x nhé mọi người
( x + 1 ) 2 = ( x + 1 ) 0 == 1
x + 1 = 1
x = 1 - 1 = 0