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![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
x-2 | 1 | -1 | 3 | -3 |
x | 3 | 1 | 5 | -1 |
b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
x-2 | 1 | -1 | 13 | -13 |
x | 3 | 1 | 15 | -11 |
c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+7 | 1 | -1 | 2 | -2 |
x | -6 | -8 | -5 | -9 |
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
Vậy: x∈{0;-7}
b) Ta có: \(\left(x+12\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
Vậy: x∈{-12;3}
c) Ta có: \(\left(-x+5\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-5\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
Vậy: x∈{3;5}
d) Ta có: \(x\left(2+x\right)\left(7-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
Vậy: x∈{-2;0;7}
e) Ta có: \(\left(x-1\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=3\end{matrix}\right.\)
Vậy: x∈{-2;1;3}
g) Ta có: \(\left(x-5\right)\left(x^2-81\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2-81=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=81\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\\x=-9\end{matrix}\right.\)
Vậy: x∈{-9;5;9}
h) Ta có: \(x^3+27=0\)
\(\Leftrightarrow x^3=-27\)
hay x=-3
Vậy: x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-3\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=12\end{cases}}\)
\(\Rightarrow x\in\left\{3;12\right\}\)
\(\left(x^2-81\right)\left(x^2+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-81=0\\x^2+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x\in\varnothing\end{cases}}\Leftrightarrow x=9\)
\(\Rightarrow x=9\)
\(\left(x-4\right)\left(x+2\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-4\\x+2\end{cases}}\)trái dấu
\(TH1:\hept{\begin{cases}x-4>0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< -2\end{cases}}\Leftrightarrow x\in\varnothing\)
\(TH2:\hept{\begin{cases}x-4< 0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 4\\x>-2\end{cases}}\Leftrightarrow x\in\left\{-1;0;1;2;3\right\}\)
Vậy \(x\in\left\{-1;0;1;2;3\right\}\)
Bài làm
a) x( 2 + x )( 7 - x ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
Vậy x = 0 hoặc x = -2 hoặc x = 7
b) ( x - 1 )( x + 2 )( x - 3 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=3\end{matrix}\right.\)
Vậy x = 1 hoặc x = -2 hoặc x = 3
c) ( x - 5 )( x2 - 81 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2-81=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=81\Leftrightarrow x=\pm9\end{matrix}\right.\)
Vậy x = 5 hoặc x = \(\pm\) 9 .
# Học tốt #
a) \(x\left(2+x\right)\left(7-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{0;-2;7\right\}\)
b) \(\left(x-1\right)\left(x+2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{1;-2;3\right\}\)
c) \(\left(x-5\right)\left(x^2-81\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x^2-81=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x^2=81\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=9\\x=-9\end{matrix}\right.\)
Vậy \(x\in\left\{5;9;-9\right\}\)