Question

Найти корень
(x ^ 2 - 6x)/(3x - 1) = (3x ^ 2 - 4x)/(1 - 3x)

обобобобоб

Answer 1

$\displaystyle \frac{x^2-6x}{3x-1}=\frac{3x^2-4x}{1-3x},x\neq \frac{1}{3}\\\\(x^2-6x)(1-3x)=(3x^2-4x)(3x-1)\\(x^2-6x)(1-3x)-(3x^2-4x)(3x-1)=0\\x^2-3x^3-6x+18x^2-(9x^3-3x^2-12x^2+4x)=0\\x^2-3x^3-6x+18x^2-(9x^3-15x^2+4x)=0\\x^2-3x^3-6x+18x^2-9x^3+15x^2-4x=0\\34x^2-12x^3-10x=0\\2x(17x-6x^2-5)=0\\2x(-6x^2+17x-5)=0\\2x(-6x^2+15x+2x-5)=0\\2x(-3x(2x-5)+2x-5)=0\\2x(2x-5)(-3x+1)=0\\x(2x-5)(-3x+1)=0\\x=0,2x-5=0,-3x+1=0\\\\x=0,x=\frac{5}{2},x=\frac{1}{3}\\\\x_1=0,x_2=\frac{5}{2}=2,5$