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Ta có: \(x^3-5x^2+6x=0\)
\(\Leftrightarrow x\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
Vậy: S={0;2;3}
\(x^3-5x^2+6x=0\)
\(\Leftrightarrow x^3-2x^2-3x^2+6x=0\)
\(\Leftrightarrow x^2\left(x-2\right)-3x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-3x\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
\(S=\left\{0,2,3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
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a: \(\Leftrightarrow4\left(2x+1\right)-3\left(6x-1\right)=2x+1\)
=>8x+4-18x+3=2x+1
=>-10x+7=2x+1
=>-12x=-6
hay x=1/2
b: \(\Leftrightarrow4x^2-12x+7x-21-x^2=3x^2+6x\)
=>5x-21=6x
=>-x=21
hay x=-21
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)(2x-1)^2-0,25=0`
`<=>(2x-1-0,5)(2x-1+0,5)=0`
`<=>(2x-1,5)(2x-0,5)=0`
`<=>[(x=0,75)(x=0,25):}`
`b)x^2+9=6x`
`<=>(x-3)^2=0`
`<=>x-3=0`
`<=>x=3`
`c)(x^2-4)-3x-6=0`
`<=>(x-2)(x+2)-3(x+2)=0`
`<=>(x+2)(x-2-3)=0`
`<=>(x+2)(x-5)=0`
`<=>[(x=-2),(x=5):}`
a: =>(2x-1-0,5)(2x-1+0,5)=0
=>(2x-1,5)(2x-0,5)=0
=>x=0,25 hoặc x=0,75
b: =>x^2-6x+9=0
=>(x-3)^2=0
=>x-3=0
=>x=3
c: =>(x-2)(x+2)-3(x+2)=0
=>(x+2)(x-5)=0
=>x=5 hoặc x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
1:
a: =>(|x|+4)(|x|-1)=0
=>|x|-1=0
=>x=1; x=-1
b: =>x^2-4>=0
=>x>=2 hoặc x<=-2
d: =>|2x+5|=2x-5
=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0
=>x=0(loại)
![](https://rs.olm.vn/images/avt/0.png?1311)
sửa đề :
\(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\Leftrightarrow x=1\)