Question

đặt nhân tử chung

 

Answer 1

h: =3(x^2+3x-10)

=3(x^2+5x-2x-10)

=3[x(x+5)-2(x+5)]

=3(x+5)(x-2)

e: =x^2-2x-4x+8

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

=(x-2)(x-4)

h: =x^2-3x-4x+12

=x(x-3)-4(x-3)

=(x-3)(x-4)

Answer 2

Lời giải:
b. $3x^2+9x-30=3(x^2+3x-10)$

$=3[(x^2-2x)+(5x-10)]=3[x(x-2)+5(x-2)]=3(x-2)(x+5)$

e. $x^2-6x+8=(x^2-2x)-(4x-8)=x(x-2)-4(x-2)=(x-4)(x-2)$

h. $x^2-7x+12=(x^2-3x)-(4x-12)=x(x-3)-4(x-3)=(x-3)(x-4)$