Question

Даю 25 баллов!! Помогите срочно с теоремой Виета!!!!!!!!!!! ​

Answer 1

Ответ:

$2392\frac{1}{3}$

Объяснение:

$x^2-5x+6=0\\\left \{ {{x1+x2=5} \atop {x1*x2=6}} \right. \\x1=2\\x2=3\\x^2-5x+6=(x-2)(x-3)\\(\frac{3}{x-3}+\frac{4}{(x-2)(x-3)}+\frac{2x}{x-2}):\frac{2x+1}{3}-\frac{x-12}{3(3-x)}+x\\\frac{3(x-2)+4+2x(x-3)}{(x-3)(x-2)}*\frac{3}{2x+1}-\frac{x-12}{3(3-x)}+x\\\frac{3x-6+4+2x^2-6x}{(x-2)(x-3)}*\frac{3}{2x+1}-\frac{x-12}{3(3-x)}+x\\\frac{2x^2-3x-2}{(x-2)(x-3)}*\frac{3}{2x+1}-\frac{x-12}{3(3-x)}+x\\2x^2-3x-2=0\\D=3^2-4*2*(-2)=9+16=25=5^2\\x1=\frac{3+5}{4}=2\\x2=\frac{3-5}{4}=-\frac{1}{2}$

$2x^2-3x-2=2(x-2)(x+\frac{1}{2})=(x-2)(2x+1)\\\frac{(x-2)(2x+1)}{(x-2)(x-3)} *\frac{3}{2x+1}-\frac{x-12}{3(3-x)}+x\\\frac{3}{x-3}-\frac{x-12}{3(3-x)}+x \\\frac{-9}{3(3-x)}-\frac{x-12}{3(3-x)}+x =\frac{-9-x+12}{3(3-x)}+x=\frac{3-x}{3(3-x)}+x\\=\frac{1}{3}+x=2392\frac{1}{3}$