Question

Hola, buenas tardes, necesito ayuda con este problema, es urgente.
Tema: Suma y resta de fracciones heterogéneas.
$\dfrac{3x + 3}{6x + 4} - \dfrac{ {x}^{2} }{6x {}^{2} + x - 2} + \dfrac{4x}{8x - 4}$
Si no sabes no respondas, gracias. ​

Answer 1

Sumas y Restas de Fracciones Algebraicas

1) Se factoriza cada polinomio

2) se saca comun denominador
3) se calcula el numerador

4) se resuelve las restas y sumas

5) se simplifica

$\bf{\dfrac{3x+3}{6x+4}- \dfrac{x^{2}}{6x^{2} +x-2}+ \dfrac{4x}{8x-4} = }\qquad Factorizamos\\ \\ \\ \bf{\dfrac{3(x+1)}{6(x+\frac{4}{6})}- \dfrac{x^{2}}{6x^{2} +x-2}+ \dfrac{4x}{4(x-1)} = }\qquad simplificamos\\ \\ \\ \bf{\dfrac{(x+1)}{2(x+\frac{2}{3})}- \dfrac{x^{2}}{6x^{2} +x-2}+ \dfrac{x}{(x-1)} = }$

Debemos factorizar el numerador que posee un trinomio

$\bf{{6x^{2} +x-2} } \ se \ factoriza \ con \ Bhaskara\\ \\ \\ x_1_y_2= \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a} \qquad a= 6\qquad b=1\qquad c= -2 \\ \\ \\ x_1_y_2= \dfrac{-1\pm\sqrt{1^{2}-4(6)(-2)}}{2(6)} \\ \\ \\ x_1_y_2= \dfrac{-1\pm\sqrt{1+48}}{2} \\ \\ \\ x_1_y_2= \dfrac{-1\pm\sqrt{49}}{12} \\ \\ \\ x_1_y_2= \dfrac{-1\pm7}{12} \\ \\ \\ x_1= \dfrac{-1+7}{12} \qquad\qquad x_2= \dfrac{-1-7}{12} \\ \\ \\ x_1= \dfrac{6}{12} \qquad\qquad\qquad x_2= \dfrac{-8}{12} \qquad SIMPLIFICAMOS$

$\bf x_1= \dfrac{1}{2} \qquad\qquad x_2= \dfrac{-2}{3} \\ \\ \\ Entonces \\ \\ \\ 6x^{2} +x-2= (x-\frac{1}{2})(x+\frac{2}{3})$
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Ahora continuamos con la Resta y la Suma
$\bf{\dfrac{3x+3}{6x+4}- \dfrac{x^{2}}{6x^{2} +x-2}+ \dfrac{4x}{8x-4} = }\qquad Factorizamos\\ \\ \\ \bf{\dfrac{3(x+1)}{6(x+\frac{4}{6})}- \dfrac{x^{2}}{6x^{2} +x-2}+ \dfrac{4x}{4(x-1)} = }\qquad simplificamos\\ \\ \\ \bf{\dfrac{(x+1)}{2(x+\frac{2}{3})}- \dfrac{x^{2}}{6x^{2} +x-2}+ \dfrac{x}{(x-1)} = }\\ \\ \\ \bf{\dfrac{(x+1)}{2(x+\frac{2}{3})}- \dfrac{x^{2}}{(x-\frac{1}{2})(x+\frac{2}{3}) }+ \dfrac{x}{(x-1)} = }\qquad Comun \ denominador$

$\bf{\dfrac{(x+1)(x-\frac{1}{2})(x-1)\ - \ 2x^{2}(x-1) +2x(x+\frac{2}{3})(x-\frac{1}{2}) }{2(x+\frac{2}{3})(x-\frac{1}{2})(x-1) } } = }\quad Resolvemos \\ \\ \\ \\ \bf{\dfrac{(x^{2}-1)(x-\frac{1}{2})\ - \ (2x^{3}-2x^{2} ) +(2x^{2} +\frac{4}{3}x)(x-\frac{1}{2}) }{2(x+\frac{2}{3})(x-\frac{1}{2})(x-1) } } = }\\ \\ \\ \\ \bf{\dfrac{(x^{3}-\frac{1}{2}x^{2} -x+\frac{1}{2})\ - \ (2x^{3}-2x^{2} ) +(2x^{3} -x^{2} +\frac{4}{3}x^{2} -\frac{2}{3}x) }{2(x+\frac{2}{3})(x-\frac{1}{2})(x-1) } } = }$

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

Espero que te sirva, salu2!!!!