Question

E(x)=(x-1/x-7-3x-1/x+7+3x²-19x/x²-49):x²+10x+25/x²-2x-35 este egal cu?

Answer 1

E(x)=[(x-1)/(x-7)-(3x-1)/(x+7)+(3x²-19x)/(x²-49)] :(x²+10x+25)/(x²-2x-35)

x²-49 =(x-7)(x+7)
x²+10x+25=(x+5)²
x²-2x-35=x²-7x+5x-35=x(x-7)+5(x-7)=(x-7)(x+5)

E(x)=[(x-1)/(x-7)-(3x-1)/(x+7)+(3x²-19x)/(x-7)(x+7)] :(x+5)²/(x-7)(x+5)=
=[(x-1)(x+7)-(3x-1)(x-7)+(3x²-19x)]/(x-7)(x+7)] :(x+5)/(x-7)=
=(x²-x+7x-7-3x²+x+21x-7+3x²-19x)/(x-7)(x+7) ×(x-7)/(x+5)=
=(x²+9x-14)/(x+7)(x+5)