Question

Представьте выражение в виде дроби:

Answer 1

1)y²/(y-5)(x+1)+2/(x+1)=(y²²+2y-10)/(y+2)(x+1)
2)2b/(a(a+2b)-b/a(2-3a)+(6b-2)/((2-3a)(a+2b)=
=(4b-6b²-ab-2b²+6ab+2a)/a(2-3a)(a+2b)=(4b-2b²-ab+2a)/a(2-3a)(a+2b)=
=[2b(2-b)+a(2-b)]/a(2-3a)(a+2b)=(2b+a)(2-b)/a(2-3a)(a+2b)=(2-b)/a(2-3a)
3)(a²+3b)/(a-b)(a+3) -(ab-3a)/(a+b)(a+3) + 1=
=(a³+a²b+3ab+3b²-a²b+ab²+3a²-3ab+a³+3a²²-ab²-3b²)/(a²-b²)(a+3)=
=(2a³+6a²)/(a²-b²)(a+3)=2a²(a+3)/(a²-b²)(a+3)==2a²/(a²-b²)

Answer 2

y²/(y-5)(x+1)+2/(x+1)=(y²+2y-10)/(y+5)(x+1)