найдите вертикальную и горизонтальную асимптоты функции f(x)=2x³-x²-2x+1/x²+3x+2

Question

Алгебра: найдите вертикальную и горизонтальную асимптоты функции f(x)=2x³-x²-2x+1/x²+3x+2

waleruwka · Accepted Answer

N4 1)=c³(ac+a-1) 3)=-((5xy+15x)-(40y+120)= -5x(y+3)-40(y+3)= (y+3)(-5x-40)=-5(y+3)(x+8) 4)= (x³-x²y)+(x²-xy)= x²(x-y)+x(x-y)= (x-y)(x²+x)= x(x-y)(x+1)
N51)= (a²-b²)-(a-b)= (a-b)(a+b)-(a-b)= (a-b)(a+b-1) 2)=(4m²-n²)+(4m-2n)=
=(2m-n)(2m+n)+2(2m-n)=(2m-n)(2m+n+2)
3)=(x²-y²)-(x+y)=(x-y)(x+y)-(x-y)= (x-y)(x+y-1)
5)=(с²-d²)+(c+d)=(c+d)(c-d)+(c+d)=(c+d)(c-d+1)
N1) 1)=9(a²-1)=9(a-1)(a+1) 2)=2m(1-m²)=2m(1-m)(1+m) 3)=x²(1-x²)=x²(1-x)(1+x)
4)= y³(1-y²)=y³(1-y)(1+y) 5)m(x²-y²)=m(x+y)(x-y)
N2) 1)=(x²-1)(x^4+x²+1)=(x-1)(x+1)(x^4+x²+1) 2)=2(x³-8)=2(x-2)(x²+2x+4)

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