\(x^4-5x^2+4=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)

`#3107.101107`

a)

`A = 2x^2 + 5x^3 + x^2y`

`= x^2 * (2 + 5x + y)`

b)

`5x(x - 1) + 15(x - 1)`

`= (5x + 15)(x - 1)`

`= 5(x + 3)(x - 1)`

\(a)A=2x^2+5x^3+x^2y\\=x^2(2+5x+y)\\=x^2(5x+y+2)\\---\\b)5x(x-1)+15(x-1)\\=(5x+15)(x-1)\\=5(x+3)(x-1)\)

\(Toru\)

1. 5x² - 4x

= x(5x - 4)

2. 8x²(x - 3y) - 12x(x - 3y)

= (8x² - 12x)(x - 3y)

= 4x(2x - 3)(x - 3y)

3. 3(x - y) - 5x(y - x)

= 3(x - y) + 5x(x - y)

= (3 + 5x)(x - y)

Đặt x²+5x+1=t chẳng hạn. Khi đó: (x²+5x+1)(x²+5x+3)-15=t.(t+2)-15=t²+2t-15. Giải phương trình bậc hai ta được: t=3 hoặc t=-5. Phương trình bậc hai có 2 nghiệm x₁, x₂ thì được viết dưới dạng nhân tử là: a(x-x₁)(x-x₂).

Vậy (x²+5x+1)(x²+5x+3)-15=(t-3)(t+5)=(x²+5x-2)(x²+5x+6). Có gì sai sót mong bạn bỏ qua cho =))

\(=x^2-5x+\dfrac{25}{4}-\dfrac{29}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{29}{4}\)

\(=\left(x-\dfrac{5}{2}-\dfrac{\sqrt{29}}{2}\right)\left(x-\dfrac{5}{2}+\dfrac{\sqrt{29}}{2}\right)\)

\(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24\\ =\left[\left(x^2+5x\right)^2-6\left(x^2+5x\right)\right]+\left[4\left(x^2+5x\right)-24\right]\\ =\left(x^2+5x\right)\left(x^2+5x-6\right)+4\left(x^2+5x-6\right)\\ =\left(x^2+5x-6\right)\left(x^2+5x+4\right)\\ =\left(x^2-x+6x-6\right)\left(x^2+4x+x+4\right)\\ =\left[x\left(x-1\right)+6\left(x-1\right)\right]+\left[x\left(x+4\right)+\left(x+4\right)\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4\right)\left(x+6\right)\)

(x2+5x)2−2(x2+5x)−24=[(x2+5x)2−6(x2+5x)]+[4(x2+5x)−24]=(x2+5x)(x2+5x−6)+4(x2+5x−6)=(x2+5x−6)(x2+5x+4)=(x2−x+6x−6)(x2+4x+x+4)=[x(x−1)+6(x−1)]+[x(x+4)+(x+4)]=(x−1)(x+1)(x+4)(x+6)