a) x¹² + 4 = x¹² + 4x⁶ + 4 - 4x⁶ = (x⁶ + 2)² - (2x³)² 

= (x⁶ - 2x³ + 2)(x⁶ + 2x³ + 2)

b) 4x⁸ + 1 = 4x⁸ + 4x⁴  + 1 - 4x⁴ = (2x⁴ + 1)² - (2x²)² 

= (2x⁴ + 2x² + 1)(2x⁴ - 2x²  + 1)

c) x⁷ + x⁵ - 1 = x⁷ - x + x⁵ + x² - (x² - x  + 1) = x(x⁶ - 1) + x²(x³ + 1) - (x² - x + 1)

= x(x³ - 1)(x³ + 1) + x²(x + 1)(x² - x + 1) - (x2 - x + 1)

= (x⁴ - x)(x + 1)(x² - x + 1) + (x³ + x²)(x² - x + 1) - (x² - x + 1)

= (x⁵ + x⁴ - x² - x + x³ + x² - 1)(x² -x + 1)

= (x⁵ + x⁴ + x³ - x - 1)(x² - x + 1)

d) x⁷ + x⁵ + 1 = x⁷ - x + x⁵ - x² + (x² + x + 1)

= x(x³ - 1)((x³ + 1) + x²(x³ - 1) + (x² + x + 1)

= (x⁴ + x)(x  - 1)(x² + x + 1) + x²(x - 1)((x² + x + 1) + (x² + x + 1)

= (x² + x + 1)(x⁵ - x⁴ + x² - x + x³ - x² + 1)

= (x² + x + 1)(x⁵ - x⁴ + x³ - x + 1)

1)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)

2) Bạn xem lại đề!

\(x^3-x^2y+3x-3y\)

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

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

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

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

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

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

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

Lời giải:

1. 
$x^3+3x^2-16x-48=(x^3+3x^2)-(16x+48)=x^2(x+3)-16(x+3)$

$=(x+3)(x^2-16)=(x+3)(x-4)(x+4)$

2.

$4x(x-3y)+12y(3y-x)=4x(x-3y)-12y(x-3y)=(x-3y)(4x-12y)=4(x-3y)(x-3y)=4(x-3y)^2$

3.

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

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

ta có :

bài a) bn trên đã dẫn link cho bn r

bài b)

Đặt x-y=a;y-z=b;z-x=c 

\(=>a+b+c=x-y+y-z+z-x=0\)

\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=a^3+b^3+c^3\)

Theo câu a)\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\) (do a+b+c=0)

\(=>a^3+b^3+c^3=3abc=>\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

a) Ta có :

\(a^3+b^3+c^3-3abc\)

\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a+b^2\right)-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

P/s tham khảo nha

hok tốt

\(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

a,\(x^5+x-1=x^5+x^4-x^2-x^4-x^3+x+x^3+x^2-1=\left(x^5+x^4-x^2\right)-\left(x^4+x^3-x\right)+\left(x^3+x^2-1\right)=x^2\left(x^3+x^2-1\right)+x\left(x^3+x^2-1\right)+\left(x^3+x^2-1\right)=\left(x^2+x+1\right)\left(x^3+x^2-1\right)\)b,\(y\left(y-2\right)-5=y^2-2y-5=\left(y^2-2y+1\right)-6=\left(y-1\right)^2-\sqrt{6^2}=\left(y-1-\sqrt{6}\right)\left(y-1+\sqrt{6}\right)\)