Question

[Minh Triều]

phân tích đa thức thành nhân tử   (x+2)(x+3)(x+4)(x+5)-24

Answer 1

[Nguyễn Thị BÍch Hậu]

<=>[(x+2)(x+5)][(x+3)(x+4)]-24=\(\left(x^2+7x+10\right)\left(\left(x^2+7x+12\right)\right)-24\)(1)

đặt x^2+7x+11=t

=> (1)<=> (t-1)(t+1)-24=t^2-1-24=t^2-25=(t-5)(t+5)

<=> \(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

Answer 2

[Dunghandsome]

(x+2)(x+3)(x+4)(x+5)=24

(x+x+x+x)(2+3+4+5)=24

(x.4)14=24

x.4=24:14

x.4=2

x=2:4

X=1/2