Question

2x^2/x^3-8+x+1/x^2+2x+4=3/x-2

Answer 1

\(\frac{2x^2}{x^3-8}+\frac{x+1}{x^2+2x+4}=\frac{3}{x-2}\\ \Leftrightarrow\frac{2x^2}{x^3-8}+\frac{\left(x+1\right)\cdot\left(x-2\right)}{x^3-8}-\frac{3\cdot\left(x^2+2x+4\right)}{x^3-8}=0\\ \Leftrightarrow\frac{2x^2}{x^3-8}+\frac{x^2-x-2}{x^3-8}-\frac{3x^2+6x+12}{x^3-8}=0\\ \Leftrightarrow\frac{2x^2+x^2-x-2-3x^2-6x-12}{x^3-8}=0\\ \Leftrightarrow\frac{-7x-14}{x^3-8}=0\\ \Rightarrow-7x-14=0\\ \Rightarrow x=-\frac{14}{7}=-2\)