Question

ДАЮ 100 БАЛЛОВ
Решить уравнение методом замены переменной.(во вложении)

Answer 1

$\frac{x^3-x+3}{x}-\frac{x}{x^3-x+3}=\frac{8x^3-8x+24}{9x}\\\\ \frac{x^3-x+3}{x}-\frac{x}{x^3-x+3}=\frac{8*(x^3-x+3)}{9*x}\\\\ t-\frac{1}{t}=\frac{8}{9}t,\ where\ t=\frac{x^3-x+3}{x}\\\\ \frac{t^2-1}{t}=\frac{8t}{9}\\\\ \frac{9(t^2-1)}{9*t}-\frac{8t*t}{9*t}=0\\\\ \frac{9t^2-9-8t^2}{9t}=0\\\\ t^2-9=0\ \ \ AND\ \ \ 9t\neq0\\\\ (t-3)*(t+3)=0\ \ \ AND\ \ \ t\neq0\\\\ (\ \ t-3=0\ \ \ OR\ \ \ t+3=0\ \ )\ \ \ AND\ \ \ t\neq0\\\\ (\ \ t=3\ \ \ OR\ \ \ t=-3\ \ )\ \ \ AND\ \ \ t\neq0$

$t=\pm3\ \ \ AND\ \ \ t\neq0\\\\ t=\pm3\\\\ \frac{x^3-x+3}{x}=-3\ \ \ OR\ \ \ \frac{x^3-x+3}{x}=3\\\\ \frac{x^3-x+3+3x}{x}=0\ \ \ OR\ \ \ \frac{x^3-x+3-3x}{x}=0\\\\ \frac{x^3+2x+3}{x}=0\ \ \ OR\ \ \ \frac{x^3-4x+3}{x}=0\\\\ (\ \ x^3+2x+3=0\ \ \ AND\ \ \ x\neq0\ \ )\ \ \ OR\ \ \ (\ \ x^3-4x+3=0\ \ \ AND\ \ \ x\neq0\ \ )\\\\ (\ \ x^3+2x+3=0\ \ \ OR\ \ \ x^3-4x+3=0\ \ )\ \ \ AND\ \ \ x\neq0\\\\ (\ \ x^3+1+2x+2=0\ \ \ OR\ \ \ x^3-1-4x+4=0\ \ )\ \ \ AND\ \ \ x\neq0$

$(\ (x+1)(x^2-x+1)+2(x+1)=0\ OR\ (x-1)(x^2+x+1)-4(x+1)=0\ )\ \ \ AND\ \ \ x\neq0\\\\ (\ (x+1)(x^2-x+1+2)=0\ OR\ (x-1)(x^2+x+1-4)=0\ )\ \ \ AND\ \ \ x\neq0\\\\ (\ (x+1)(x^2-x+3)=0\ OR\ (x-1)(x^2+x-5)=0\ )\ \ \ AND\ \ \ x\neq0\\\\ (x+1)(x^2-x+3)=0\ OR\ (x-1)(x^2+x-5)=0\\\\ (\ x=-1\ \ OR\ \ x^2-x+3=0\ )\ \ OR\ \ (\ x=1\ \ OR\ \ x^2+x-5=0\ )\\\\ x=\pm1\ \ OR\ \ x^2-x+3=0\ \ OR\ \ x^2+x-5=0$

///////////////////

$x^2-x+3>0\ \ for\ \ any\ \ real\ \ x,\ \ according\ \ to\ \ the\ \ fact\ \ that\\\\D=(-1)^2-4*3<0$

///////////////////

$x=\pm1\ \ OR\ \ x^2+x-5=0\\\\ x=\pm1\ \ OR\ \ x=\frac{-1\pm\sqrt{(-1)^2-4*(-5)}}{2}\\\\ x=\pm1\ \ OR\ \ x=\frac{-1\pm\sqrt{21}}{2}$

Answer:$x\in\{\pm1;\ \frac{-1\pm\sqrt{21}}{2}\}$