Question

помогите сделать последний пример

Answer 1

$(frac{x+1}{3x-3}+frac{6}{3x^{2}-3}-frac{x+3}{3*(x+3)})*frac{3x^{2}-3}{4}=frac{x+4}{2}$

1) $frac{x+1}{3*(x-1)}+frac{6}{3*(x-1)(x+1)}-frac{x+3}{3*(x+3)}=frac{(x+1)^{2}(x+3)+6(x+3)-(x+3)(x^{2}-1)}{3*(x-1)*(x+1)(x+3)}=$$frac{(x^{2}+2x+1)(x+3)+6x+18-(x^{3}-x+3x^{2}-3)}{3*(x-1)*(x+1)(x+3)}=$$frac{x^{3}+2x^{2}+x+3x^{2}+6x+3+6x+18-x^{3}+x-3x^{2}+3}{3*(x-1)*(x+1)(x+3)}=frac{2x^{2}+14x+24}{3*(x-1)*(x+1)(x+3)}$$=frac{2(x^{2}+7x+12)}{3*(x-1)*(x+1)(x+3)}=$$frac{2(x+3)(x+4)}{3*(x-1)*(x+1)(x+3)}=frac{2(x+4)}{3*(x-1)*(x+1)}$

2) $frac{2(x+4)}{3*(x-1)*(x+1)}*frac{3(x-1)(x+1)}{4}=frac{x+4}{2}$