Giải làm gì cho nó mệt 

\(x^2+5x-2=\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}\right)-\frac{25}{4}-2=\left(x+\frac{5}{2}\right)^2-\frac{33}{4}\)

\(=\left(x+\frac{5}{2}\right)^2-\left(\frac{\sqrt{33}}{2}\right)^2=\left(x+\frac{5}{2}-\frac{\sqrt{33}}{2}\right)\left(x+\frac{5}{2}+\frac{\sqrt{33}}{2}\right)\)

\(=\left(x+\frac{5-\sqrt{33}}{2}\right)\left(x+\frac{5+\sqrt{33}}{2}\right)\)

x²-8x+15

=x²-3x-5x+15

=x.(x-3)-5.(x-3)

=(x-3)(x-5)

a) \(x^3+4x^2-21x\)

\(=x\left(x^2+4x-21\right)\)

\(=x\left(x^2-3x+7x-21\right)\)

\(=x\left[x\left(x-3\right)+7\left(x-3\right)\right]\)

\(=x\left(x-3\right)\left(x+7\right)\)

b) \(5x^3+6x^2+x\)

\(=x\left(5x^2+6x+1\right)\)

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

\(=x\left[5x\left(x+1\right)+\left(x+1\right)\right]\)

\(=x\left(x+1\right)\left(5x+1\right)\)

c) \(x^3-7x+6\)

\(=x^3+2x^2-3x-2x^2-4x+6\)

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

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

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

d) \(3x^3+2x-5\)

\(=3x^3+3x^2+5x-3x^2-3x-5\)

\(=x\left(3x^2+3x+5\right)-\left(3x^2+3x+5\right)\)

\(=\left(x-1\right)\left(3x^2+3x+5\right)\)

x⁴-3x²+1=x⁴-2x²+1-x²

=(x²-1)²-x²

=(x²-1-x)(x²-1+x)

\(x^4-5x^2y^2+4y^4\)

\(=\left(x^2\right)^2-2x^22y^2+\left(2y^2\right)^2-x^2y^2\)

\(=\left(x^2-2y^2\right)^2-\left(xy\right)^2\)

\(=\left(x^2-2y^2-xy\right)\left(x^2-2y^2+xy\right)\)

\(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\\ x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\\ 45+x^3-5x^2-9x=x^2\left(x-5\right)-9\left(x-5\right)=\left(x-3\right)\left(x+3\right)\left(x-5\right)\)

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

   =(3x-5)(x-y)

Bn j ơi gõ latex đi bn !

Ta nhắc lại: Phương trình bậc hai phân tích được thành nhân tử khi và chỉ khi nó tồn tại nghiệm.

Ta thấy: `x^2-4x+12=(x-2)^2+8>=8>0AAx` nên ta không thể phân tích nhân tử cho phương trình này.

x² - 4x - 12

= x² + 2x - 6x - 12

= (x² + 2x) - (6x + 12)

= x(x + 2) - 6(x + 2)

= (x + 2)(x - 6) 

`x^2-4x+12`

`=x^2+2x-6x+12`

`=x(x+2)-6(x+2)`

`=(x+2)(x-6)`

Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]       
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
 

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

2: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)

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

5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)

6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)