Question

докажите торжество
$\frac{2 {a}^{2} - 7a + 3 }{ {a}^{3} - 1} - \frac{1 - 2a}{ {a}^{2} + a + 1} - \frac{3}{a - 1} = \frac{1}{a - 1}$
пж срочно!!!!!!!!!!!!!!!!​

Answer 1

$\frac{2 {a}^{2} + 7a + 3 }{ {a}^{3} - 1 } - \frac{1 - 2a}{ {a}^{2} + a + 1 } - \frac{3}{a - 1} = \\ = \frac{2 {a}^{2} + 7a + 3 }{ ({a}^{} - 1)( {a}^{2} + a + 1) } - \frac{1 - 2a}{ {a}^{2} + a + 1 } - \frac{3}{a - 1} = \\ = \frac{2 {a}^{2} + 7a + 3 - (1 - 2a)(a - 1) - 3( {a}^{2} + a + 1) }{(a - 1)( {a}^{2} + a + 1) } = \\ = \frac{2 {a}^{2} + 7a + 3 - a + 1 + 2 {a}^{2} - 2a - 3 {a}^{2} - 3a - 3 }{(a - 1)( {a}^{2} + a + 1) } = \\ = \frac{ {a}^{2} + a + 1 }{(a - 1)( {a}^{2} + a + 1)} = \frac{1}{a - 1}$