Question

Necesito que alguien me pase las ecuaciones ya lista para ser operadas, osea simplificadas

Answer 1

Explicación paso a paso:

5.

$x(x - 1) + 1 = \frac{5}{6} + \frac{x(2x - 1)}{3}$

${x}^{2} - x + 1 = \frac{5}{6} + \frac{2 {x}^{2} - x }{3}$

$mcm = 6$

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

$6 {x}^{2} - 6x + 6 = 5 + 4 {x}^{2} - 2x$

$2 {x}^{2} - 4x + 1 = 0$

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

$x = \frac{4 + - 2\sqrt{2} }{4}$

$x = \frac{2 + - \sqrt{2} }{2}$

6.

$11(x - 1 {)}^{2} = (2x - 3 {)}^{2} + 4 {x}^{2} + 1$

$11 {x}^{2} - 22x + 11 = 4 {x}^{2} - 12x + 9 + 4 {x}^{2} + 1$

$3 {x}^{2} - 10x + 1 = 0$

$x = \frac{ - ( - 10) + - \sqrt{( - 10 {)}^{2} - 4(3)(1)} }{2(3)}$

$x = \frac{10 + - 2 \sqrt{22} }{6}$

$x = \frac{5 + - \sqrt{22} }{3}$

7.

$(3x - \frac{1}{2} )(3x + \frac{1}{2} ) - 2x = 8 {x}^{2} - 1$

$9 {x}^{2} - 2x - \frac{1}{4} = 8 {x}^{2} - 1$

${x}^{2} - 2x + \frac{3}{4} = 0$

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

$x = \frac{2 + - 1}{2}$

$x1 = \frac{2 + 1}{2} = \frac{3}{2}$

$x2 = \frac{2 - 1}{2} = \frac{1}{2}$

8.

$(x + 1)( \frac{3}{2} - 2(1 - x)) = 3 {x}^{2} + \frac{11(x - 1)}{2}$

$2 {x}^{2} + \frac{3x - 1}{2} = 3 {x}^{2} + \frac{11x - 11}{2}$

$mcm = 2$

$2(2 {x}^{2} + \frac{3x - 1}{2} = 3 {x}^{2} + \frac{11x - 11}{2} )$

$4 {x}^{2} + 3x - 1 = 6 {x}^{2} + 11x - 11$

$- 2 {x}^{2} - 8x + 10 = 0$

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

$x = \frac{8 + - 12}{ - 4}$

$x1 = \frac{8 + 12}{ - 4} = - 5$

$x2 = \frac{8 - 12}{ - 4} = 1$

9.

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

$6 {x}^{2} + 6x - 6 = 4 {x}^{2} + x - 3$

$2 {x}^{2} + 5x - 3 = 0$

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

$x = \frac{ - 5 + - 7}{4}$

$x1 = \frac{ - 5 + 7}{4} = \frac{1}{2}$

$x2 = \frac{ - 5 - 7}{4} = - 3$

10.

$3x + 2 = \frac{3(4x + 3)}{4 - 3x}$

$3x + 2 = \frac{12x + 9}{4 - 3x}$

$- 9 {x}^{2} + 6x + 8 = 12x + 9$

$- 9 {x}^{2} - 6x - 1 = 0$

$x = \frac{ - ( - 6) + - \sqrt{( - 6 {)}^{2} - 4( - 9)( - 1)} }{2( - 9)}$

$x = \frac{6 + - 0}{ - 18}$

$x1 = \frac{6 + 0}{ - 18} = - \frac{1}{3}$

$x2 = \frac{6 - 0}{ - 18} = - \frac{1}{3}$

11.

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

$\frac{6x + 4}{x - 1} = \frac{3x + 4}{x + 2}$

$6 {x}^{2} + 16x + 8 = 3 {x}^{2} + x - 4$

$3 {x}^{2} + 15x + 12 = 0$

$x = \frac{ - (15) + - \sqrt{(15 {)}^{2} - 4(3)(12)} }{2(3)}$

$x = \frac{ - 15 + - 9}{6}$

$x1 = \frac{ - 15 + 9}{6} = - 1$

$x2 = \frac{ - 15 - 9}{6} = - 4$

12.

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

$mcm = 20$

$4(x + 2 {)}^{2} - 5( {x}^{2} - 9) = 10(x + 3 {)}^{2} + 4$

$- {x}^{2} + 16x + 61 = 10 {x}^{2} + 60x + 94$

$- 11 {x}^{2} - 44x - 33 = 0$

$x = \frac{ - ( - 44) + - \sqrt{( - 44 {)}^{2} - 4( - 11)( - 33) } }{2( - 11)}$

$x = \frac{44 + - 22}{ - 22}$

$x1 = \frac{44 + 22}{ - 22} = - 3$

$x2 = \frac{44 - 22}{ - 22} = - 1$

psdt: si te ayudé sígueme y ponme como mejor respuesta :)