1. Phân tích các đa thức sau thành nhân tử:
`b) 2x (x - 3) - 3y (3 - x)`
`=(x-3)(2x-3y)`
`c) 81 - (5x - 3)^2`
`=[9-(5x-3)][9+(5x-3)]`
`=(12-5x)(6+5x)`
`d) (x - 3) . (x + 3) - (x - 3)^2`
`=(x-3)[(x+3)-(x-3)]`
`=(x-3)9`
2. Tìm x, biết:
`a) x^3 - 25x = 0`
`x(x^2-25)=0`
`x(x-5)(x+5)=0`
\(\left[{}\begin{matrix}x=0\\x-5=0\Rightarrow x=5\\x+5=0\Rightarrow x=-5\end{matrix}\right.\)
`b) 3x (x - 7) - x + 7 = 0`
`3x(x-7)-(x-7)=0`
`(x-7)(3x-1)=0`
\(\left[{}\begin{matrix}x-7=0\Rightarrow x=7\\3x-1=0\Rightarrow x=\dfrac{1}{3}\end{matrix}\right.\)
`c) x. (x - 2 ) - (x - 3 )^2 = 0`
có sai ko nhỉ?
`d) x^3- x^2+ 8x - 8 = 0`
`x^2(x-1)+8(x-1)=0`
`(x^2 + 8) (x-1)=0`
\(\left[{}\begin{matrix}x^2+8=0\Rightarrow x\in\varnothing\\x-1=0\Rightarrow x=1\end{matrix}\right.\)
` 2x (x - 3) - 3y (3 - x)`
`= 2x (x - 3) + 3y (x-3)`
`=(x-3)(2x+3y)`
`---------`
`81 - (5x - 3)^2`
`=9^2 - (5x - 3)^2`
`=(9- 5x+3)(9x-5x-3)`
`-----------`
`(x - 3) . (x + 3) - (x - 3)^2`
`=x^2 -9 - x^2 - 6x + 9`
`= -6x`
`-----------`
`x^3 - 25x=0`
`<=> x(x^2-25)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm5\end{matrix}\right.\)
`-------------`
`3x (x - 7) - x + 7 = 0`
`<=> 3x (x - 7) -(x-7)=0`
`<=> (x-7)(3x-1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\3x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0+7\\3x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{1}{3}\end{matrix}\right.\)
`------------`
` x. (x - 2 ) - (x - 3 )^2 = 0`
`<=> x^2 - 2x -(x^2 - 6x + 9)=0`
`<=> x^2 - 2x -x^2+6x-9=0`
`<=>4x-9=0`
`<=>4x=9`
`<=>x=9/4`
`------------`
Bài 2:
a: =>x(x^2-25)=0
=>x(x-5)(x+5)=0
hay \(x\in\left\{0;5;-5\right\}\)
b: =>(x-7)(3x-1)=0
=>x=1/3 hoặc x=7
c: =>x^2-2x-x^2+6x-9=0
=>4x-9=0
=>x=9/4
d: =>x^2(x-1)+8(x-1)=0
=>x-1=0
=>x=1
