a: \(x^4+3x^3+x^2+3x\)

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

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

c: \(x^2-xy-x+y\)

\(=x\left(x-y\right)-\left(x-y\right)\)

\(=\left(x-y\right)\left(x-1\right)\)

a) x²+x-2
= x²-x+2x-2
= x(x-1)+2(x-1)
= (x+2)(x-1)
b) 2x²+5x+3
= 2x²+2x+3x+3
= 2x(x+1)+3(x+1)
= (2x+3)(x+1)
c) 3x²+5x-2
= 3x²+6x-1x-2
= 3x(x+2)-1(x+2)
= (3x-1)(x+2)

\(=\left(x^2+4x-3\right)^2-5\left(x^2+4x-3\right)+6x^2\)

\(=x^4+16x^2+9+8x^3-24x-6x^2-5x^2-20x+15+6x^2\)

\(=x^4+8x^3+11x^2-44x+24\)

\(=\left(x^4-x^3\right)+\left(9x^3-9x^2\right)+\left(20x^2-20x\right)-\left(24x-24\right)\)

\(=x^3\left(x-1\right)+9x^2\left(x-1\right)+20x\left(x-1\right)-24\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+9x^2+20x-24\right)\)

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

b/ \(=y^4+\left(-2x^2-34\right)y^2+32xy+x^4-34x^2+225\)

câu này bn coi lại đc k , mk k lm ra 

\(A=\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)

\(A=\left[\left(x+1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x+2\right)\right]+4x^2\)

\(A=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

Đặt \(p=x^2-4,5x-8\)ta có :

\(A=\left(p-2,5x\right)\left(p+2,5x\right)+4x^2\)

\(A=p^2-\left(2,5x\right)^2+4x^2\)

\(A=p^2-6,25x^2+4x^2\)

\(A=p^2-2,25x^2\)

\(A=p^2-\left(1,5x\right)^2\)

\(A=\left(p-1,5x\right)\left(p+1,5x\right)\)

Thay \(p=x^2-4,5x-8\)vào A ta có :

\(A=\left(x^2-4,5x-8-1,5x\right)\left(x^2-4,5x-8+1,5x\right)\)

\(A=\left(x^2-6x-8\right)\left(x^2-3x-8\right)\)

\(\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)

\(=\left(x+1\right)\left(x-8\right)\left(x-4\right)\left(x+2\right)+4x^2\)

\(=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

  Đặt \(x^2-2x-8=t\)

  Ta có : \(\left(t-5x\right)t+4x^2\)

\(=t^2-5xt+4x^2\)

\(=t^2-2.\frac{5}{2}xt+\frac{25}{4}x^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}\right)^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}-\frac{3}{2}x\right)\left(t-\frac{5}{2}+\frac{3}{2}x\right)\)

    Học tốt ~~

a: \(=x\left(x^2+2xy+y^2-4\right)\)

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

\(a,x^2+2x^2y+xy^2-4x=x\left(x^2+2xy+y^2-4\right)=x\left[\left(x+y\right)^2-4\right]=x\left(x+y-2\right)\left(x+y+2\right)\\ b,x^2-7xy+10y^2=\left(x^2-2xy\right)-\left(5xy-10y^2\right)=x\left(x-2y\right)-5y\left(x-2y\right)=\left(x-2y\right)\left(x-5y\right)\)

\(x-y=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\)

đúng ,mình k 2 nhé

\(x^3-2x^2-19x+20\)

\(=x^3+3x^2-4x-5x^2-15x+20\)

\(=\left(x^3+3x^2-4x\right)-\left(5x^2+15x-20\right)\)

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

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

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

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

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

x^3-2x^2-19x+20

=x^3-5x^2+3x^2-15x-4x+20

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

=(x+4)(x-1)(x-5)

Có: x³ - 2x² - 19x + 20

=> x³ - 2x² + x - 20x + 20

=> x(x² - 2x + 1) - 20(x - 1)

=> x(x - 1)² - 20(x - 1)

=> (x - 1)[x(x - 1) - 20]

=> (x - 1)(x² - x - 20)

=> (x - 1)(x² - 5x + 4x - 20)

=> (x - 1)[x(x - 5) + 4(x - 5)]

=> (x - 1)(x + 4)(x - 5)