\(x^4+4\)

= \(\left(x^2+2\right)^2-4x^2\)

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

Đa thức này không phân tích được thành nhân tử bạn nhé. 

`x^4+x^2 y^2+y^4`

`=x^4+2x^2 y^2 +y^4-x^2 y^2`

`=(x^2+y^2)^2-(xy)^2`

`=(x^2-xy+y^2)(x^2+xy+y^2)`

x⁴ - 2x³-2x² -2x -3

=(x⁴+x³)-(3x³+3x²)+(x²+x)-(3x+3)

=x³(x+1)-3x²(x+1)+x(x+1)-3(x+1)

= (x³-3x²+x-3)(x+1)

= ((x³-3x²)+(x-3))(x+1)

= (x²(x-3)+(x-3))(x+1)

=(x²+1)(x-3)(x+1)

x⁴ + 64 = (x⁴ +  16x² + 64) - 16x² = (x² + 8)² - (4x)² = (x² - 4x + 8).(x² + 4x + 8)

Ta có

x⁴ + 64

= (x⁴ +  16x² + 64) - 16x² 

= (x² + 8)² - (4x)² 

= (x² - 4x + 8).(x² + 4x + 8)

hok tốt

\(27x^3-\dfrac{1}{64}\)

\(=\left(3x\right)^3-\left(\dfrac{1}{4}\right)^3\)

\(=\left(3x-\dfrac{1}{4}\right)\left(9x^2+\dfrac{3x}{4}+\dfrac{1}{16}\right)\)

x^4+64

=(x^2)^2+8^2+2.x^2.8-2.x^2.8

=(x^2+8)^2-16x^2

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

x³+64=x³+4³=(x+4)(x²-4x+4²)=(x+4)(x²-4x+16)

\(x^3+64=x^3+4^3=(x+4)(x^2-3x+16)\)

Ta có: 
x^4 + 64 = (x²)² + 8² + 2x².8 - 2.x².8 
= (x² + 8)² - (4x)² 
= (x² - 4x + 8)(x² + 4x + 8) 

Ta có: 
x^4 + 64 = (x²)² + 8² + 2x².8 - 2.x².8 
= (x² + 8)² - (4x)² 
= (x² - 4x + 8)(x² + 4x + 8)