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a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(x+3\right)\left(5-x\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(c,x+17⋮x+3\\ x+3+14⋮x+3\\ 14⋮x+3\\ x+3\inƯ\left(14\right)=\left\{\pm14;\pm7\pm2;\pm1\right\}\)
Từ đó bạn tìm những giá trị của x nha!
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a, |x+1| =5 => x+1=+5hay -5 => x= -4 ;4 mà x>0 => x =4
b,|x-3|=7=> x-3 = -7 ; +7
=> x = -4 ; 10 mà x < 3
=> x = -4
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>xy=-18
=>x,y khác dấu
mà x<y<0
nên không có giá trị nào của x và y thỏa mãn yêu cầu đề bài
b: =>(x+1)(y-2)=3
\(\Leftrightarrow\left(x+1,y-2\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(0;5\right);\left(2;3\right);\left(-2;-1\right);\left(-4;1\right)\right\}\)
c: \(\Leftrightarrow8x-4=3x-9\)
=>5x=-5
hay x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: -7<x<-1
mà \(x\in Z\)
nên \(x\in\left\{-6;-5;-4;-3;-2\right\}\)
Vậy: \(x\in\left\{-6;-5;-4;-3;-2\right\}\)
b) Ta có: -3<x<3
mà \(x\in Z\)
nên \(x\in\left\{2;1;0;1;2\right\}\)
Vậy: \(x\in\left\{2;1;0;1;2\right\}\)
c) Ta có: \(-1\le x\le6\)
mà \(x\in Z\)
nên \(x\in\left\{-1;0;1;2;3;4;5;6\right\}\)
Vậy: \(x\in\left\{-1;0;1;2;3;4;5;6\right\}\)
d) Ta có: \(-5\le x< 6\)
mà \(x\in Z\)
nên \(x\in\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
Vậy: \(x\in\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
a) I x + 1 I = 5
=> \(\orbr{\begin{cases}x+1=5\\x+1=-5\end{cases}}\)=> \(\orbr{\begin{cases}x=5-1\\x=\left(-5\right)-1\end{cases}}\)=>\(\orbr{\begin{cases}x=4\\x=-6\end{cases}}\)
Mà a thuộc Z => x = - 6
Vậy x = - 6