a) $ 4x^4 -37x^2+9 $

Đặt $t=x^2, t \geq 0$

Ta có:

$4t^2-37t+9$

$=4t^2-t-36t+9$

$=t(4t-1)-9(4t-1)$

$=(4t-1)(t-9)$

Vậy

$ 4x^4 -37x^2+9$

$=(4x^2-1)(x^2-9)$

$=(2x-1)(2x+1)(x-3)(x+3). $

b) $ (x-y)^2 +4x-4y -12=(x-y)^2+4(x-y)-12$

Đặt $t=x-y$

Ta có:

$(x-y)^2+4(x-y)-12$

$=t^2+4t-12$

$=t^2-2t+6t-12$

$=t(t-2)+6(t-2)$

$=(t-2)(t+6)$

Vậy

$ (x-y)^2 +4x-4y -12$

$=(x-y)^2+4(x-y)-12$

$=(x-y-2)(x-y+6).$

c)  $ (x^2 + 3x)^2 + 7x^2 +21x +10 =(x^2 + 3x)^2+7(x^2 + 3x)+10 $

Đặt $t=x^2 + 3x$

Ta có:

$t^2+7t+10$

$=t^2+2t+5t+10$

$=t(t+2)+5(t+2)$

$=(t+2)(t+5)$

Vậy

$ (x^2 + 3x)^2 + 7x^2 +21x +10$

$=(x^2 + 3x)^2+7(x^2 + 3x)+10$

$=(x^2 + 3x+2)(x^2 + 3x+5).  $

a) $(x+1)(x+2)(x+3)(x+4) – 24$

$=(x+1)(x+4)(x+2)(x+3)-24$

$=(x^2+5x+4)(x^2+5x+6)-24$

Đặt $t=x^2+5x+5$

Suy ra

$(x^2+5x+4)(x^2+5x+6)-24$

$=(t-1)(t+1)-24$

$=t^2-1-24$

$=t^2-25=(t-5)(t+5)$

Vậy $(x^2+5x+4)(x^2+5x+6)-24$

$=(x^2+5x+5-5)(x^2+5x+5+5)$

$=(x^2+5x)(x^2+5x+10)$

$=x(x+5)(x^2+5x+10)$

b)  $ (x+2)(x+4)(x+6)(x+8)+16$

$=(x+2)(x+8)(x+4)(x+6)+16$

$=(x^2+10x+16)(x^2+10x+24)+16 $

Đặt $t=x^2+10x+20$

Suy ra

$(x^2+10x+16)(x^2+10x+24)+16$

$=(t-4)(t+4)+16$

$=t^2-16+16=t^2 $

Vậy

$(x^2+10x+16)(x^2+10x+24)+16$

$=(x^2+10x+20)^2 $
c)$ (x^2 + 6x +8)(x^2+8x+15) -24$

$=(x+2)(x+4)(x+3)(x+5)-24$

$=(x+2)(x+5)(x+3)(x+4)-24$

$=(x^2+7x+10)(x^2+7x+12)-24 $

Đặt $t=x^2+7x+11$

Suy ra

$(x^2+7x+10)(x^2+7x+12)-24$

$=(t-1)(t+1)-24$

$=t^2-1-24$

$=t^2-25$

$=(t-5)(t+5)$

Vậy

$(x^2+7x+10)(x^2+7x+12)-24$

$=(x^2+7x+11-5)(x^2+7x+11+5)$

$=(x^2+7x+6)(x^2+7x+16)$

a)  $(x^2 + 1)^2 + 3x(x^2+1) +2x^2$

Đặt $ t=x^2+1$, ta được:

$t^2+3xt+2x^2$

$=(t^2+xt)+(2xt+2x^2)$

$=t(t+x)+2x(t+x)$

$=(t+x)(t+2x)$

Vậy

$ (x^2 + 1)^2 + 3x(x^2+1) +2x^2$

$=(x^2+1+x)(x^2+1+2x)$.

b)  $ (x^2 +4x +8)^2 +3x(x^2 + 4x+ 8) + 2x^2. $

Đặt $ t=x^2+4x+8$, ta được:

$t^2+3xt+2x^2$

$=(t^2+xt)+(2xt+2x^2)$

$=t(t+x)+2x(t+x)$

$=(x+t)(t+2x)$

Vậy

$ (x^2 +4x +8)^2 +3x(x^2 + 4x+ 8) + 2x^2$

$=(x+x^2+4x+8)(x^2+4x+8+2x)$

$=(x^2+5x+8)(x^2+6x+8)$.
c)  $ 4(x^2 +x +1)^2 + 5x(x^2 + x + 1)+ x^2. $

Đặt $ t=x^2+x+1$, ta được:

$4t^2+5xt+x^2$

$=(4t^2+4xt)+(xt+x^2)$

$=4t(t+x)+x(t+x)$

$=(x+t)(4t+x)$

Vậy

$ 4(x^2 +x +1)^2 + 5x(x^2 + x + 1)+ x^2$

$=(x^2 +x +1+x)[4(x^2 +x +1)+x]$

$=(x+1)^2(4x^2+5x+4)$.

a)  $ x^4 + 6x^3 +11x^2 + 6x+1$

$=x^2\left(x^2+6x+11+\dfrac{6}{x}+\dfrac{1}{x^2}\right)$

$=x^2\left[\left(x^2+\dfrac{1}{x^2}\right)+6\left(x+\dfrac{1}{x}\right)+11\right]$

Đặt $t=x+\dfrac{1}{x} \Rightarrow t^2=\left(x+\dfrac{1}{x}\right)^2 \Rightarrow x^2+\dfrac{1}{x^2}=t^2-2$

$x^2\left[\left(x^2+\dfrac{1}{x^2}\right)+6\left(x+\dfrac{1}{x}\right)+11\right]$

$=x^2(t^2-2+6t+11)$

$=x^2(t^2+6t+9)$

$=x^2(t+3)^2$

$=x^2\left(x+\dfrac{1}{x}+3\right)^2.$
b) $ x^4 + 5x^3 -12x^2 + 5x +1. $

$=x^2\left(x^2+5x-12+\dfrac{5}{x}+\dfrac{1}{x^2}\right)$

$=x^2\left[\left(x^2+\dfrac{1}{x^2}\right)+5\left(x+\dfrac{1}{x}\right)-12\right]$

Đặt $t=x+\dfrac{1}{x} \Rightarrow t^2=\left(x+\dfrac{1}{x}\right)^2 \Rightarrow x^2+\dfrac{1}{x^2}=t^2-2$

$=x^2\left[\left(x^2+\dfrac{1}{x^2}\right)+5\left(x+\dfrac{1}{x}\right)-12\right]$

$=x^2(t^2-2+5t-12)$

$=x^2(t^2+5t-14)$

$=x^2(t^2-2t+7t-14)$

$=x^2[t(t-2)+7(t-2)]$

$=x^2(t-2)(t+7)$

$=x^2\left(x+\dfrac{1}{x}-2\right)\left(x+\dfrac{1}{x}+7\right).$

c) $ 6x^4 + 5x^3 -38x^2 + 5x+ 6. $

$=x^2\left(6x^2+5x-38+\dfrac{5}{x}+\dfrac{6}{x^2}\right)$

$=x^2\left[6\left(x^2+\dfrac{1}{x^2}\right)+5\left(x+\dfrac{1}{x}\right)-38\right]$

Đặt $t=x+\dfrac{1}{x} \Rightarrow t^2=\left(x+\dfrac{1}{x}\right)^2 \Rightarrow x^2+\dfrac{1}{x^2}=t^2-2$

$=x^2\left[6\left(x^2+\dfrac{1}{x^2}\right)+5\left(x+\dfrac{1}{x}\right)-38\right]$

$=x^2[6(t^2-2)+5t-38]$

$=x^2(6t^2-12+5t-38)$

$=x^2(6t^2+5t-50)$

$=x^2(6t^2-15t+20t-50)$

$=x^2(2t-5)(3t+10)$

$=x^2\left[2\left(x+\dfrac{1}{x}\right)-5\right]\left[3\left(x+\dfrac{1}{x}\right)+10\right].$

a) $ (x+1)(x-4)(x+2)(x-8) + 4x^2. $

$=(x+1)(x-8)(x-4)(x+2)+4x^2$

$=(x^2-7x-8)(x^2-2x-8)+4x^2$

Đặt $t=x^2-8 $

$(x^2-7x-8)(x^2-2x-8)+4x^2$

$=(t-7x)(t-2x)+4x^2$

$=t^2-9xt+14x^2+4x^2$

$=t^2-9xt+18x^2$

$=t^2-3xt-6xt+18x^2$

$=t(t-3x)-6x(t-3x)$

$=(t-3x)(t-6x)$

$=(x^2-8-3x)(x^2-8-6x).$
b)  $ (x-1)(x+2)(x+3)(x-6)+32x^2. $

$=(x-1)(x-6)(x+2)(x+3)+32x^2$

$=(x^2-7x+6)(x^2+5x+6)+32x^2$

Đặt $t=x^2+6 $

$(x^2-7x+6)(x^2+5x+6)+32x^2$

$=(t-7x)(t+5x)+32x^2$

$=t^2-2xt-35x^2+32x^2$

$=t^2-2xt-3x^2$

$=t^2+xt-3xt-3x^2$

$=t(t+x)-3x(t+x)$

$=(t+x)(t-3x)$

$=(x^2+6+x)(x^2+6-3x).$

c) $ (x+2)(x-4)(x+6)(x-12) +36x^2. $

$=(x+2)(x-12)(x-4)(x+6)+36x^2$

$=(x^2-10x-24)(x^2+2x-24)+36x^2$

Đặt $t=x^2-24 $

$=(x^2-10x-24)(x^2+2x-24)+36x^2$

$=(t-10x)(t+2x)+36x^2$

$=t^2-8xt-20x^2+36x^2$

$=t^2-8xt+16x^2$

$=(t-4x)^2$

$=(tx^2-24-4x)^2$.