x.(x+3)=0
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a.
\(x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b.
\(\left(x+12\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
c.
\(\left(-x+5\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
d.
Em kiểm tra lại ngoặc cuối của câu này
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-3\right)^3+\left(x+3\right)^3=0\)
\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)
\(\Leftrightarrow x^2\left(2x+54\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)
\(b,\left(x+1\right)^3-\left(x-1\right)^3=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)
\(\Leftrightarrow6x^2=-2\)
\(\Leftrightarrow x^2=-3\) ( vô lí)
Vậy pt vô nghiệm
\(c,x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(d,4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+2-x-3\right)\left(x+2+x+3\right)=0\)
\(\Leftrightarrow-\left(2x+5\right)=0\)
\(\Leftrightarrow-2x-5=0\)
\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)
Học tốt nha you <3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-3\right)^3+\left(x+3\right)^3=0\)
\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)
\(\Leftrightarrow x^2\left(2x+54\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)
\(b,\left(x+1\right)^3-\left(x-1\right)^3=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)
\(\Leftrightarrow6x^2=-2\)
\(\Leftrightarrow x^2=-3\) ( vô lí)
Vậy pt vô nghiệm
\(c,x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(d,4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+2-x-3\right)\left(x+2+x+3\right)=0\)
\(\Leftrightarrow-\left(2x+5\right)=0\)
\(\Leftrightarrow-2x-5=0\)
\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)
Học tốt nha you <3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-3\right)^3+\left(x+3\right)^3=0\)
\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)
\(\Leftrightarrow x^2\left(2x+54\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)
\(b,\left(x+1\right)^3-\left(x-1\right)^3=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)
\(\Leftrightarrow6x^2=-2\)
\(\Leftrightarrow x^2=-3\) ( vô lí)
Vậy pt vô nghiệm
\(c,x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(d,4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+2-x-3\right)\left(x+2+x+3\right)=0\)
\(\Leftrightarrow-\left(2x+5\right)=0\)
\(\Leftrightarrow-2x-5=0\)
\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)
Học tốt nha you <3
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
Làm theo công thức: tích bằng 0 thì một trong x thừa số bằng 0 rồi xét các trường hợp
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,x.\left(x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}}\)
\(2,\left(x+12\right).\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+12=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-12\\x=3\end{cases}}}\)
\(3,\left(-x+5\right).\left(3-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x+5=0\\3-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}}\)
4/ \(x.\left(2+x\right).\left(7-x\right)=0\)
\(\hept{\begin{cases}x=0\\2+x=0\\7-x=0\end{cases}}\) => \(\hept{\begin{cases}x=0\\x=-2\\x=7\end{cases}}\)
Vậy \(x=\left\{0,-2,7\right\}\)
5/ \(\left(x-1\right).\left(x+2\right).\left(-x-3\right)=0\)
\(\hept{\begin{cases}x-1=0\\x+2=0\\-x-3=0\end{cases}}\)=> \(\hept{\begin{cases}x=1\\x=-2\\x=-3\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x=0 hoặc x+7=0
suy ra x=0 hoặc x=-7
b) x+12=0 hoặc x-3=0
x=-12 hoặc x=3
c) x=0 hoặc x+2=0 hoặc 7-x=0
x=0 hoặc x=-2 hoặc x=7
d) x-1=0 hoặc x+2=0 hoặc -x-3=0
suy ra x=1 hoặc x=-2 hoặc x=-3
Bài làm
x( x + 7 ) = 0
<=> x = 0 hoẵ x + 7 = 0
=> x = 0 hoặc x = -7
Vậy x = 0 hoặc x = -7
( x + 12 )( x - 3 ) = 0
<=> x + 12 = 0 hoặc x - 3 = 0
=> x = -12 hoặc x = 3
Vậy x = -12 hoặc x = 3
( -x + 5 )( 3 - x ) = 0
<=> -x + 5 = 0 hoặc 3 - x = 0
=> x = 5 hoặc x = 3
Vậy x = 5 hoặc x = 3
x( 2 + x )( 7 - x ) = 0
<=> x = 0 hoặc 2 + x = 0 hoặc 7 - x = 0
=> x = 0 hoặc x = -2 hoặc x = 7
Vậy x = 0 hoặc x = -2 hoặc x j 7
( x - 1 )( x + 2 )( -x - 3 ) = 0
<=> ( x - 1 ) = 0 hoặc x + 2 = 0 hoặc ( -x - 3 ) = 0
<=> x = 1 hoăc x = -2 hoặc x = ( -3)
Vậy x = 1 hoặc x = 2 hoặc x = -3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(x+5\right)\left(x-4\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0.\\x-4=0.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5.\\x=4.\end{matrix}\right.\)
Vậy..........
\(b,\left(x-1\right)\left(x-3\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\x-3=0.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=3.\end{matrix}\right.\)
Vậy..........
\(c,\) Sửa đề:
\(\left(3-x\right)\left(x-3\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0.\\x-3=0.\end{matrix}\right.\)
\(\Leftrightarrow x=3.\)
Vậy..........
\(d,x\left(x+1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0.\\x+1=0.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0.\\x=-1.\end{matrix}\right.\)
Vậy..........
![](https://rs.olm.vn/images/avt/0.png?1311)
(x+12).(x-3)=0
=> ta có 2 trường hợp:
+) x+12 = 0 => x = 0-12=-12
+) x-3 = 0 => x = 0+3=3
Vậy x= 0 hoặc x = 3
(-x+5).(3-x)=0
=> Ta có 2 trường hợp:
+) -x +5= 0 => -x=-5 => x= 5
+) 3 - x =0 => x=3-0=3
Vậy x= 5 hoặc 3
x . (x + 3) = 0
để tích là 0 thì 1 trong 2 thừa số là 0
=> x = 0
x . (x + 3) = 0
=> x = 0
Bài này nghĩ z thui