Question

Помогите с алгеброй.

Answer 1

здравствуй.

$(\frac{3}{x-3} +\frac{4}{x^2-5x+6} +\frac{2x}{x-2} ):\frac{2x+1}{3} -\frac{x-12}{9-3x} +x=2358.3\\\\1)\ \frac{3}{x-3} +\frac{4}{x^2-5x+6} +\frac{2x}{x-2}=\frac{3}{x-3} +\frac{4}{(x-2)(x-3)} +\frac{2x}{x-2}=\\\\=\frac{(x-2)*3+4+2x*(x-3)}{(x-2)(x-3)} =\frac{3x-6+4+2x^2-6x}{(x-2)(x-3)}=\frac{(x-2)(1+2x)}{(x-2)(x-3)}=\frac{1+2x}{x-3}\\\\2)\ \frac{1+2x}{x-3}:\frac{2x+1}{3}=\frac{(1+2x)3}{(x-3)(2x+1)}=\frac{3}{x-3}$

$3)\ \frac{3}{x-3}-\frac{x-12}{9-3x}=\frac{3}{x-3}-\frac{x-12}{3(3-x)}=-\frac{3}{3-x}-\frac{x-12}{3(3-x)}=\\\\=\frac{-9-x+12}{3(3-x)}=\frac{3-x}{3(3-x)}=\frac{1}{3} \\\\4)\ \frac{1}{3} +x=\frac{1}{3}(1+3x)=\frac{3x+1}{3}$

при x = 2358

$\frac{3*2358+1}{3} =\frac{7075}{3} =2358\frac{1}{3} =2358.3$

Ответ: 2358.3

удачи в учебе.