Đáp án+Giải thích các bước giải:

1)

a)

(x-12)(x+7)+2x+14

=(x-12)(x+7)+(2x+14)

=(x-12)(x+7)+2(x+7)

=(x-12)(x+7)+2(x+7)

=(x+7)(x-12+2)

=(x+7)(x-10)

b)

$x^2$-6x+8

=($x^2$-2x)-(4x-8)

=x(x-2)-4(x-2)

=(x-2)(x-4)

c)

$9x^2$+9xy-(x+y)

=9x(x+y)-(x+y)

=(x+y)(9x-1)

d)

($x^2$-2xy+$y^2$)-81

=$(x-y)^2$-$9^2$

=(x-y-9)(x-y+9)

e)

$(x+4)^2$-15(4+x)

=$(x+4)^2$-15(x+4)

=(x+4)(x+4-15)

=(x+4)(x-11)

2)

a)

2x(x-2019)-x+2019=0

⇔2x(x-2019)-(x-2019)=0

⇔(x-2019)(2x-1)=0

⇔\(\left[ \begin{array}{l}x-2019=0\\2x-1=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=2019\\x=\frac{1}{2}\end{array} \right.\) 

Vậy S={2019,`(1)/(2)`}

b)

$x^3$-25x=0

⇔x($x^2$-25)=0

⇔\(\left[ \begin{array}{l}x=0\\x^2-25=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=0\\x=±5\end{array} \right.\) 

Vậy S={0,±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 S={2,-2}

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`