Question

Решите и докажите,что тождественно равные выражение$\frac{1}{x-2} +\frac{x-2}{x^2+2x+4} -\frac{6x}{x^3-8} \neq \frac{2x-4}{x^2+2x+4}$

Answer 1

$\frac{1}{x - 2 } + \frac{x - 2}{ {x}^{2} + 2x + 4 } - \frac{6x}{ {x}^{3} - 8} = \frac{1}{x - 2 } + \frac{x - 2}{ {x}^{2} + 2x + 4 } - \frac{6x}{ (x - 2)( {x}^{2} + 2x + 4) } = \frac{ {x}^{2} + 2x + 4 + {(x - 2)}^{2} - 6x}{ (x - 2)( {x}^{2} + 2x + 4) } = \frac{ {x}^{2} + 2x + 4 + {x}^{2} - 4x + 4 - 6x}{ (x - 2)( {x}^{2} + 2x + 4) } = \frac{ 2 {x}^{2} - 8x + 8 }{ (x - 2)( {x}^{2} + 2x + 4) } = \frac{ 2 {(x - 2)}^{2} }{ (x - 2)( {x}^{2} + 2x + 4) } = \frac{ 2x - 4}{ {x}^{2} + 2x + 4 }$