Solution:
[tex](x+1)(2x-4)(\frac{1}{x}+1)=(x+1)(2x-4)(1-\frac{5}{2x}-4)\\(x+1)(2x-4)(\frac{1}{x}+1)-(x+1)(2x-4)(1-\frac{5}{2x}-4) = 0\\(x+1)(2x-4)(\frac{1}{x}+1-(1-\frac{5}{2x}-4))=0\\(x+1)(2x-4)(\frac{1}{x}+1-(-3\frac{5}{2x}))=0\\(x+1)(2x-4)(\frac{1}{x}+1+3+\frac{5}{2x})=0\\(x+1)(2x-4)(\frac{1}{x}+4+\frac{5}{2x})=0\\(x+1)(2x-4)(\frac{5x^{2}+8x+2}{2x} )=0\\[/tex]
[tex]x+1=0\\x=-1\\\\2x-4=0\\x=2\\\\\frac{5x^{2}+8x+2}{2x}=0\\x=\frac{-4+\sqrt[]{6}}{5}\\x=\frac{-4-\sqrt[]{6}}{5}[/tex]
Answer:
[tex]x=-1\\x=2\\x=\frac{-4+\sqrt[]{6}}{5}\\x=\frac{-4-\sqrt[]{6}}{5}[/tex]
Hope this was helpful.
