Đáp án:

 Bài 1 :

a)  $ax$ - $2x$ - $a^{2}$ + $2a$

= ( $ax$ - $2x$ ) - ( $a^{2}$ + $2a$ )

= $x$ ( $a$ - $2$ ) - $a$ ( $a$ - $2$ ) 

= ( $x$ - $a$ ) . ( $a$ - $2$ )

b) $x^{3}$ - $2x^{2}y$ + $xy^{2}$ - $9x$ 

 = $x$ $( x - y )^{2}$ - $3^{2}$

= $x$ $( x - y $ - $3$ ). $( x - y $ + $3$ ) 

Xin ctlhn kiếm điểm cho nhóm ạ !

Bài 3 : 

a) $\frac{x - 18}{3-2x}$ + $\frac{11x}{3-2x}$

= $\frac{x-18+11x}{3x-2}$ = $\frac{18-12x}{3x-2}$

b) $\frac{4x}{x+2}$ - $\frac{3x}{x-2}$ + $\frac{12x}{x^{2}-4}$

= $\frac{4x.(x-2)}{(x+2).(x-2)}$ - $\frac{3x(x+2)}{(x-2).(x+2)}$ + $\frac{12x}{(x-2).(x+2)}$

= $\frac{4x.(x-2)-3x(x+2)+ 12x }{(x-2).(x+2)}$

= $\frac{13x}{1}$ = $13x$

Câu 1:

a) ax - 2x - a² + 2a

= x . (a-2) - a . (a-2)

= (x-a) . (a-2) 

 b) x³ - 2x²y + xy² - 9x

= x . ( x² - 2xy + y² - 9 )

= x . [ (x + y)² - 3² ]

= x. (x+y-3). ( x+y+3) 

Câu 2:

b) 5x(x-3) - 2x+6 = 0

⇔ 5x(x-3)- 2(x-3) = 0

⇔ (5x-2)(x-3) = 0

⇔ 5x-2 = 0 hoặc x-3=0

+) 5x -2 = 0 ⇔ 5x = 2 ⇔ x = 2/5

+) x - 3 = 0 ⇔ x = 3

Vậy x ∈ { 2/5;3} 

Câu 3:

${\frac{x-18}{3-2x} + \frac{11x}{3-2x} }$

${ = \frac{ x-18+11x}{3-2x}}$

${=\frac{12x-18}{3-2x}}$

${=\frac{-6.(3-2x)}{3-2x}}$

${=-6}$

${b) \frac{4x}{x+2} - \frac{3x}{x-2}+ \frac{12x}{x^{2}-4} }$

${= \frac{ 4x.(x-2)}{(x+2).(x-2)} - \frac {3x.(x+2)}{(x-2).(x+2)} + \frac{12x}{(x-2).(x+2)} }$

$ {= \frac { 4x.(x-2) - 3x.(x+2) + 12x}{(x-2).(x+2)} }$

${=\frac { 4x^{2} -8x -3x^{2} -6x +12x}{(x-2).(x+2)}}$

${=\frac{x^{2}-2x}{(x-2).(x+2)}}$

$ {=\frac {x.(x-2)}{(x-2).(x+2)}}$

${=\frac {x}{x+2}}$