Question

Разложите на множител:
16x^4-81
x^2-x-y^2-y

Answer 1

$16x^4-81=(4x^2)^2-9^2=(4x^2-9)(4x^2+9)= \\ =((2x)^2-3^2)(4x^2+9)=(2x-3)(2x+3)(4x^2+9) \\ \\ x^2-x-y^2-y=(x^2-y^2)-(x+y)= \\ =(x-y)(x+y)-1(x+y)=(x+y)(x-y-1)$