Question

Phân tích các đt sau thành nhân tử 

1)(x^2+x)^2+4(x^2+x)-12

2)(x^2+x)^2+9x^2+9x+14

3)(x^2+5x)^2+10x^2+50x+24

Answer 1

1/ Đặt x²+x=t

=>(x²+x)²+4(x²+x)-12=t²+4t-12=t²+6t-2t-12=t(t+6)-2(t+6)=(t-2)(t+6)=(x²+x-2)(x²+x+6)=(x²-x+2x-2)(x²+x+6)

=[x(x-1)+2(x-1)](x²+x+6)=(x-1)(x+2)(x²+x+6)

2/ Đặt x²+x=t 

=>(x²+x)²+9x²+9x+14=(x²+x)²+9(x²+x)+14=t²+9t+14=t²+2t+7t+14=t(t+2)+7(t+2)=(t+2)(t+7)=(x²+x+2)(x²+x+7)

3/ Đặt x²+5x=t

=>(x²+5x)²+10x²+50x+24=(x²+5x)²+10(x²+5x)+24=t²+10t+24=t²+4t+6t+24=t(t+4)+6(t+4)=(t+4)(t+6)=(x²+5x+4)(x²+5x+6)

=(x²+x+4x+4)(x²+2x+3x+6)=[x(x+1)+4(x+1][x(x+2)+3(x+2)]=(x+1)(x+4)(x+2)(x+3)=(x+1)(x+2)(x+3)(x+4)