Question

Найти сумму корней или корень уравнения

Answer 1

$\frac{x(3-x)}{\frac1{x-7}-\frac2{x-10}}=\frac4{\frac2{x-10}+\frac1{7-x}}\\\\O.D.3.:\\\begin{cases}\frac1{x-7}-\frac2{x-10}\neq0\\\frac2{x-10}+\frac1{7-x}\neq0\\x-7\neq0\\x-10\neq0\\7-x\neq0\end{cases}\Rightarrow\begin{cases}\frac{x-10-2x+14}{(x-7)(x-10)}\neq0\\\frac{14-2x+x-10}{(x-10)(7-x)}\neq0\\x\neq7\\x\neq10\end{cases}\Rightarrow\begin{cases}x\neq4\\x\neq7\\x\neq10\end{cases}$

$\frac{3x-x^2}{\frac1{x-7}-\frac2{x-10}}-\frac4{\frac2{x-10}+\frac1{7-x}}=0\\\\\frac{3x-x^2}{\frac1{x-7}-\frac2{x-10}}-\frac4{\frac2{x-10}-\frac1{x-7}}=0\\\\\frac{3x-x^2}{\frac1{x-7}-\frac2{x-10}}+\frac4{\frac1{x-7}-\frac2{x-10}}=0\\\\\frac{-x^2+3x+4}{\frac1{x-7}-\frac2{x-10}}=0\\\\-x^2+3x+4=0\\x^2-3x-4=0\\D=9-4\cdot(-4)=9+16=25\\x_{1,2}=\frac{3\pm5}2\\x_1=-1;\\x_2=4\;-\;He\;nogx.\;no\;O.D.3.$