Đáp án:

Giải thích các bước giải:

a) \(x^4+16\)

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

\(=\left(x^2\right)^2+2x^2.4+4^2-2x^2.4\)

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

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

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

b) \(x^4y^4+64=x^4y^4+16x^2y^2+64-16x^2y^2=\left(x^2y^2+8\right)^2-16x^2y^2=\left(x^2y^2-4xy+8\right)\left(x^2y^2+4xy+8\right)\)

c) `x^4y^4 + 4`

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

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

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

f) \(x^8+x+1=x^8-x^2+\left(x^2+x+1\right)=x^2\left(x^6-1\right)+\left(x^2+x+1\right)=x^2\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)=x^2\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)+x^2+x+1=\left(x^2+x+1\right)\text{[}x^2\left(x+1\right)\left(x-1\right)\left(x^2-x+1\right)+1\text{]}\)

g) x⁸ + x⁷ + 1

= x⁸ + x⁷ - x⁶ + x⁶ + 1

= x⁶( x² + x +1) - [ ( x³)² - 1]

= x⁶( x² + x +1) - ( x³ - 1)( x³ + 1)

= x⁶( x² + x +1) - ( x -1)( x² + x +1)( x³ + 1)

= ( x² + x +1)[x⁶ - ( x -1)( x³ + 1)]

= ( x² + x +1)[ x⁶ - ( x⁴ + x - x³ - 1)

= ( x² + x +1)( x⁶ - x⁴ - x + x³ - 1)

\(k)x^4+4y^4\)

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

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

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

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

l) `a^2-b^2-2x(a-b)`

`= (a-b)(a+b)-2x(a-b)`

`=(a-b)(a+b-2x)`

m) `a^2-b^2-2x(a+b)`

`= (a-b)(a+b)-2x(a+b)`

`=(a+b)(a-b-2x)`