\(\left(x+2\right)\left(x^2-3x+5\right)=\left(x+2\right)x^2\\\left(x+2\right)\left(x^2-3x+5\right)-\left(x+2\right)x^2=0\\ \left(x+2\right)\left(5-3x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)
\(\dfrac{-7x^2+4}{x^3+1}=\dfrac{5}{x^2-x+1}-\dfrac{1}{x+1}\\ \dfrac{-7x^2+4}{x^3+1}=\dfrac{5\left(x+1\right)-\left(x^2-x+1\right)}{x^3+1}\\ \Rightarrow-7x^2+4=-x^2+6x-4\\ 6x^2+6x-8=0\\ x^2+x-\dfrac{4}{3}=0\\ x^2+x+\dfrac{1}{4}=\dfrac{4}{3}+\dfrac{1}{4}\\ \left(x+\dfrac{1}{2}\right)^2=\dfrac{19}{12}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\sqrt{\dfrac{19}{12}}\\x+\dfrac{1}{2}=-\sqrt{\dfrac{19}{12}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{12}}-\dfrac{1}{2}\\x=-\sqrt{\dfrac{19}{12}}-\dfrac{1}{2}\end{matrix}\right.\)
\(2x^2-x=3-6x\\ 2x^2+5x-3=0\\ x^2+\dfrac{5}{2}x-\dfrac{3}{2}=0\\ x^2+2.\dfrac{5}{4}x+\dfrac{25}{16}=\dfrac{49}{16}\\ \left(x+\dfrac{5}{4}\right)^2=\dfrac{49}{16}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{4}=\dfrac{7}{4}\\x+\dfrac{5}{4}=-\dfrac{7}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}-\dfrac{5}{4}=\dfrac{1}{2}\\x=-\dfrac{7}{4}-\dfrac{5}{4}=-3\end{matrix}\right.\)
ĐKXĐ: x khác +-2
\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \dfrac{x^2-4x+4-3x-6}{x^2-4}=\dfrac{2x-22}{x^2-4}\\ \Rightarrow x^2-12x-2=2x-22\\ x^2-14x+20=0\\ x^2-14x+49=49-20\\ \left(x-7\right)^2=29\\ \Rightarrow\left[{}\begin{matrix}x-7=29\\x-7=-29\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=36\\x=-22\end{matrix}\right.\)
