Question

$E(x)= \frac{x^4+x^2+1}{(x+1)^2-x}$
Va rog sa-mi explicati

Answer 1

$E(x) = \frac{ {x}^{4} + {x}^{2} + 1 }{ {(x + 1)}^{2} - x } \\ \\ E(x) = \frac{ {x}^{4} + {x}^{2} + 1 + {x}^{2} - {x}^{2} }{ {(x}^{2} + 2x + 1) - x } \\ \\ E(x) = \frac{ {x}^{4} + {x}^{2} + {x}^{2} + 1 - {x}^{2} }{ {x}^{2} + 2x + 1 - x } \\ \\ E(x) = \frac{ {x}^{4} + {2x}^{2} + 1 - {x}^{2} }{ {x}^{2} + 2x - x + 1 } \\ \\ E(x) = \frac{ { {(x}^{2} + 1) }^{2} - {x}^{2} }{ {x}^{2} + x + 1 } \\ \\ E(x) = \frac{( {(x}^{2} + 1) - x)( ({x}^{2} + 1) + x) }{ {x}^{2} + x + 1 } \\ \\ E(x) = \frac{ {(x}^{2} + 1 - x)( {x}^{2} + 1 + x) }{ {x}^{2} + x + 1 } \\ \\ E(x) = \frac{( {x}^{2} - x + 1)( {x}^{2} + x + 1) }{ {x}^{2} + x + 1 } \\ \\ E(x) = {x}^{2} - x + 1$