Question

ejercicio resolver porfa​

Answer 1

Hola!

Respuesta:

x = 2

Explicación paso a paso:

$\frac{(a + b)x}{a - b} + \frac{ax}{a + b} - \frac{a - b}{a + b} = \frac{ax}{a - b} + \frac{ {(a + b)}^{2} }{ {a}^{2} - {b}^{2} } \\ \frac{x{(a + b)}^{2} + ax(a - b)}{ {a}^{2} - {b}^{2} } - \frac{ {(a + b)}^{2} }{ {a}^{2} - {b}^{2} } = \frac{ax}{a - b} + \frac{a - b}{a + b} \\ \frac{x {(a + b)}^{2} + ax(a - b) - {(a + b)}^{2} }{ {a}^{2} - {b}^{2} } = \frac{ax(a + b) + {(a - b)}^{2} }{ {a}^{2} - {b}^{2} } \\ {(a + b)}^{2} (x - 1) + ax(a - b) - ax(a + b) - {(a - b)}^{2} = 0 \\ (a + b)((a + b)(x - 1) - ax) + (a - b)(ax - (a - b)) = 0 \\ (a + b)(ax - a + bx - b - ax) + (a - b)(ax - a + b) = 0 \\ (a + b)(bx - a - b) + ( {a}^{2} x - {a}^{2} + ab - abx + ab - {b}^{2} ) = 0 \\ abx - {a}^{2} - ab + {b}^{2} x - ab - {b}^{2} + {a}^{2} x - {a}^{2} + ab - abx + ab - {b}^{2} ) = 0 \\ - 2 {a}^{2} + {b}^{2} x - 2 {b}^{2} + {a}^{2} x = 0 \\ x( {a}^{2} + {b}^{2} ) = 2( {a}^{2} + {b}^{2} ) \\ x = 2$