Bài 1:
`a)` Sửa đề:`(x-12)(x+7)-2x+14`
`→(x-12)(x+7)-2x-14`
`(x-12)(x+7)-2x-14`
`=(x-12)(x+7)-2(x+7)`
`=(x+7)(x-12-2)`
`=(x+7)(x-14)`
Hoặc:Sửa đề:`(x-12)(x+7)-2x+14`
`→(x-12)(x+7)+2x+14`
`(x-12)(x+7)+2x+14`
`=(x-12)(x+7)+2(x+7)`
`=(x+7)(x-12+2)`
`=(x+7)(x-10)`
`b)x²-6x+8`
`=x²-2x-4x+8`
`=x(x-2)-4(x-2)`
`=(x-2)(x-4)`
`c)9x²+9xy-(x+y)`
`=9x(x+y)-(x+y)`
`=(x+y)(9x-1)`
`d)(x²-2xy+y²)-81`
`=(x-y)²-9²`
`=(x-y+9)(x-y-9)`
`e)`Sửa đề:`(x+4)²-15(4-x)`
`→(x-4)²-15(4-x)`
`(x-4)²-15(4-x)`
`=(x-4)²+15(x-4)`
`=(x-4)(x-4+15)`
`=(x-4)(x+11)`
Hoặc:Sửa đề:`(x+4)²-15(4-x)`
`→(x+4)²-15(4+x)`
`(x+4)²-15(4+x)`
`=(x+4)²-15(x+4)`
`=(x+4)(x+4-15)`
`=(x+4)(x-11)`
Bài 2:
`a)2x(x-2019)-x+2019=0`
`⇔2x(x-2019)-(x-2019)=0`
`⇔(x-2019)(2x-1)=0`
`⇔`\(\left[ \begin{array}{l}x-2019=0\\2x-1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=2019\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy `x=2019` hoặc `x=1/2`
`b)x³-25x=0`
`⇔x(x²-25)=0`
`⇔x(x²-5²)=0`
`⇔x(x+5)(x-5)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\x+5=0\\x-5=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=-5\\x=5\end{array} \right.\)
Vậy `x=0` hoặc `x=-5` hoặc `x=5`
`c)(x+2)²-4(x+2)=0`
`⇔(x+2)(x+2-4)=0`
`⇔(x+2)(x-2)=0`
`⇔`\(\left[ \begin{array}{l}x+2=0\\x-2=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-2\\x=2\end{array} \right.\)
Vậy `x=-2` hoặc `x=2`
