Question

3 supra x la a treia + 1 plus(a doua fracție) 2 supra x la a doua minus unu​

Answer 1

Răspuns:

Bună,

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

$2) \frac{3a}{6ax ^{2} } - \frac{x}{6a ^{2}x } + \frac{a}{3 {a}^{2} {x}^{2} } = \\ \frac{1}{2 {x}^{2} } - \frac{1}{6 {a}^{2} } + \frac{3}{3a {x}^{2} } = \\ \orange{ \frac{3a ^{2} - {x}^{2} + 2a}{6 {a}^{2} {x}^{2} } }$

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

$4) \frac{2 - 3x}{ {x}^{2} - x} + \frac{2}{ x - 1 } + \frac{5}{3x} = \\ \frac{2 - 3x}{x( x - 1)} + \frac{2}{x - 1} + \frac{5}{3x} = \\ \frac{3(2 - 3x) + 3x \times 2 + 5(x - 1)}{3x(x - 1)} = \\ \frac{6 - 9x + 6x + 5x - 5}{3x(x - 1)} = \\ \orange{ \frac{2x + 1}{3x(x - 1)} }$

Sper că te-am ajutat.