Question

Решите 4 пожалуйста!!!!


Умоляю!!!

Answer 1

$\frac{2x^{2}+7x+9 }{x^{3}-1 }+\frac{4x+3}{x^{2}+x+1 }-\frac{5}{x-1}=\frac{2x^{2}+7x+9 }{(x-1)(x^{2}+x+1)}+\frac{4x+3}{x^{2}+x+1 }-\frac{5}{x-1}=\frac{2x^{2}+7x+9+4x^{2}-4x+3x-3-5x^{2}-5x-5}{(x-1)(x^{2}+x+1)}=\frac{x^{2}+x+1 }{(x-1)(x^{2} +x+1)}=\frac{1}{x-1}\\\\\frac{1}{x-1}=\frac{1}{1,1-1}=\frac{1}{0,1}=10\\\\Otvet:\boxed{10}$

Answer 2

$x=1,1\\\\\frac{2x^2+7x+9}{x^3-1}+\frac{4x+3}{x^2+x+1}-\frac{5}{x-1}=\frac{2x^2+7x+9+(4x+3)(x-1)-5(x^2+x+1)}{(x-1)(x^2+x+1)}=\\\\=\frac{2x^2+7x+9+4x^2-4x+3x-3-5x^2-5x-5}{(x-1)(x^2+x+1)}=\frac{x^2+x+1}{(x-1)(x^2+x+1)}=\frac{1}{x-1}=\\\\=\frac{1}{1,1-1}=\frac{1}{0,1}=10$