Bài 1. Phân tích các đa thức sau thành nhân tử:
A = $a^{2}$ – ab + bc – ca
B = $x^{2}$ – (a + b)x + a.b
C = $3x^{3}$ – $2x^{2}$ + 5x + 2
D = $27x^{3}$ – $27x^{2}$ + 18x – 4
E = $3x^{3}$ − $7x^{2}$ +17x − 5
F = $x^{2}$ − x − 2001.2002
G = $2x^{3}$ − $5x^{2}$ + 8x − 3
H = $3x^{3}$ −$14x^{2}$ + 4x + 3
Giúp mk với ạ, một hai câu cx đc. mk cảm ơn
Đáp án:
Giải thích các bước giải:A = (a^2 – ca) – ( ab -bc)
=a . (a -c ) – b .( a – c )
= ( a – c ) . ( a -b )
B = x^2 – ax + bx + ab
= ( x^2 – ax ) – ( bx – ab )
= x ( x – a ) – b ( x – a )
= ( x – a ) . ( x – b)
$A=a^2-ab+bc-ca$
$=a(a-b)+c(b-a)$
$=a(a-b)-c(a-b)$
$=(a-b)(a-c)$
$B=x^2-(a+b)x+a.b$
$=x^2-ax-bx+ab$
$=x(x-a)-b(x-a)$
$=(x-a)(x-b)$
$C=3x^3-2x^2+5x+2$
$=3x^3+x^2-3x^2-x+6x+2$
$=x^2(3x+1)-x(3x+1)+2(3x+1)$
$=(3x+1)(x^2-x+2)$
$D=27x^3-27x^2+18x-4$
$=27x^3-9x^2-18x^2+6x+12x-4$
$=9x^2(3x-1)-6x(3x-1)+4(3x-1)$
$=(3x-1)(9x^2-6x+4)$
$E=3x^3-7x^2+17x-5$
$=3x^3-x^2-6x^2+2x+15x-5$
$=x^2(3x-1)-2x(3x-1)+5(3x-1)$
$=(3x-1)(x^2-2x+5)$
$F=x^2-x-2001.2002$
$=x^2+2001x-2002x-2001.2002$
$=x(x+2001)-2002(x+2001)$
$=(x+2001)(x-2002)$
$G=2x^3-5x^2+8x-3$
$=2x^3-x^2-4x^2+2x+6x-3$
$=x^2(2x-1)-2x(2x-1)+3(2x-1)$
$=(2x-1)(x^2-2x+3)$
$H=3x^3-14x^2+4x+3$
$=3x^3+x^2-15x^2-5x+9x+3$
$=x^2(3x+1)-5x(3x+1)+3(3x+1)$
$=(3x+1)(x^2-5x+3)$