Question

phân tích đa thức thành nhân tử

1/(x+2)(x+3)(x+4)(x+5)-24

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

3/(x^2+x+1)(x^2+x+2)-12

4/(a^2-4)(a^2+6a+5)

Answer 1

1/(x+2)(x+3)(x+4)(x+5)-24

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

=(x+2)(x-2+7)(x+3)(x-3+7)

=[(x+2)(x-2)+7x+14][(x+3)(x-3)+7x+21]

=(x²-4+7x+14)(x²-9+7x+21)

=(x²+10+7x)(x²+12+7x)

2/(x²+x)²+4(x²+x)-12

=(x²+x)²+4(x²+x)+2²-16

=(x²+x+2)²-4²

=(x²+x+2+4)(x²+x+2-4)

=(x²+x+6)(x²+x-2)

3/(x²+x+1)(x²+x+2)-12

=(x²+x+1)(x²+x+-1+3)-12

=(x²+x+1)(x²+x+-1)+3(x²+x+1)-12

=(x²+x)-1+3(x²+x)+3-12

=(x²+x)(x²+x+3)-10

làm đến đây thì mk bí, bạn giúp suy nghĩ nốt nha

4/nó là nhân tử sẵn rồi mà

 

Answer 2

\(3/\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

\(=\left(x^2+x+1\right)\left(x^2+x+1+1\right)-12\)

\(=\left(x^2+x+1\right)^2+x^2+x+1-12\)

\(=\left(x^2+x+1\right)^2+4\left(x^2+x+1\right)-3\left(x^2+x+1\right)-12\)

\(=\left(x^2+x+1\right)\left(x^2+x+1+4\right)-3\left(x^2+x+1+4\right)\)

\(=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)

\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)