Question

toate exercitiile va rog am lipsit la clasa si imi este foarte greu sa inteleg,trebuie sa le fac si pe nota

Answer 1

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

$b) \frac{2x + 1}{x - 2} + \frac{1 - 2x}{x + 2} + \frac{ {x}^{2} + 16 }{ {x}^{2} - 4 } = \frac{2x + 1}{x - 2} + \frac{1 - 2x}{x + 2} + \frac{ {x}^{2} + 16 }{ {x}^{2} - {2}^{2} } = \frac{2x + 1}{x - 2} + \frac{1 - 2x}{x + 2} + \frac{ {x}^{2} + 16 }{(x + 2)(x - 2)} = \frac{(2x + 1)(x + 2) + (1 - 2x)(x - 2) + {x}^{2} + 16 }{(x - 2)(x + 2)} = \frac{ {2x}^{2} + 4x + x + 2 + x - 2 - {2x}^{2} + 4x + {x}^{2} + 16 }{(x - 2)(x + 2)} = \frac{( {2x}^{2} - {2x}^{2} + {x}^{2}) + (4x + x + x + 4x) + (2 - 2 + 16) }{(x - 2)(x + 2)} = \frac{ {x}^{2} + 10x + 16 }{(x - 2)(x + 2)} = \frac{(x + 2)(x + 8)}{(x - 2)(x + 2)} = \frac{x + 8}{x - 2}$

$c) \frac{x - 2}{x + 1} + \frac{x + 3}{x + 2} + \frac{5 - {x}^{2} }{ {x}^{2} + 3x + 2 } = \frac{x - 2}{x + 1} + \frac{x + 3}{x + 2} + \frac{5 - {x}^{2} }{(x + 1)(x + 2)} = \frac{(x - 2)(x + 2) + (x + 3)(x + 1) + 5 - {x}^{2} }{(x + 1)(x + 2)} = \frac{ {x}^{2} - {2}^{2} + {x}^{2} + x + 3x + 3 + 5 - {x}^{2} }{(x + 1)(x + 2)} = \frac{ {x}^{2} - 4 + {x}^{2} + x + 3x + 3 + 5 - {x}^{2} }{(x + 1)(x + 2)} = \frac{( {x}^{2} + {x}^{2} - {x}^{2}) + ( - 4 + 3 + 5) + (x + 3x) }{(x + 1)(x + 2)} = \frac{ {x}^{2} + 4 + 4x }{(x + 1)(x + 2)} = \frac{ {x}^{2} + 4x + 4 }{(x + 1)(x + 2)} = \frac{ {(x + 2)}^{2} }{(x + 1)(x + 2)} = \frac{x + 2}{x + 1}$