Question

[shoppe pi pi pi pi]

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

[Phong Thần]

1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+11=a\), ta có:

\(=\left(a+1\right)\left(a-1\right)-24\)

\(=a^2-1-24\)

\(=a^2-25\)

\(=\left(a-5\right)\left(a+5\right)\)

\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x^2+6x+x+6\right)\left(x^2+7x+16\right)\)

\(=\left[x\left(x+6\right)+\left(x+6\right)\right]\left(x^2+7x+16\right)\)

\(=\left(x+6\right)\left(x+1\right)\left(x^2+7x+16\right)\)

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

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

\(=\left(x^2+x+2\right)^2-16\)

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

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

\(=\left(x^2+x-2\right)\left(x^2+x+6\right)\)

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

Đặt \(x^2+x+1=a\), ta được

\(=a\left(a+1\right)-12\)

\(=a^2+a-12\)

\(=a^2+2.a.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}-12\)

\(=\left(a+\dfrac{1}{2}\right)^2-\dfrac{49}{4}\)

\(=\left(a+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2\)

\(=\left(a+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(a+\dfrac{1}{2}+\dfrac{7}{2}\right)\)

\(=\left(a-3\right)\left(a+4\right)\)

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

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

4) \(\left(a^2-4\right)\left(a^2+6a+5\right)\)

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

\(=\left(a-2\right)\left(a+2\right)\left[a\left(a+5\right)+\left(a+5\right)\right]\)

\(=\left(a-2\right)\left(a+2\right)\left(a+5\right)\left(a+1\right)\)