(2(x^2-16))/((x-4)(x+4))-5(x^2+x+1)(x-1)/((x^3-1) )+(x+7)^2/((x^3+21x^2+147x+343) )=2.... Question from @medinamalilo99ossnjr

medinamalilo99ossnjr @medinamalilo99ossnjr

November 2018 1 11 Report

(2(x^2-16))/((x-4)(x+4))-5(x^2+x+1)(x-1)/((x^3-1) )+(x+7)^2/((x^3+21x^2+147x+343) )=2

Klyffor $\frac{2(x^2-16)}{(x+4)(x-4)}- \frac{5(x^2+x+1)(x-1)}{x^3-1} + \frac{(x+7)^2}{x^3+21x^2+147x+343} =2$

$Lo$ $hare$ $por$ $partes$

$\frac{2(x^2-16)}{(x+4)(x-4)} \\ \\ \frac{2(x^2-16)}{x^2-16} \\ \\2$

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

$\frac{(x+7)^2}{x^3+21x^2+147x+343}$

$Usare$ $Division$ $Sintetica$ $para$ $el$ $denominador$

x³ + 21x² + 147x + 343

x³       x²     x    TI
1      21    147   343    | -7 ------>(x + 7)
         -7     -98   -343   |
________________ |____
1      14     49      0

(x + 7)(x² + 14x + 49) --------->Factorizo la ecuacion cuadratica

(x + 7)(x + 7)(x + 7)

(x + 7)³ ------------>Asi queda el denominador

$\frac{(x+7)^2}{(x + 7)^3} \\ \\ \frac{1}{x+7}$

Me queda de esta manera entonces:

$2-5+ \frac{1}{x+7} =2 
\\ \\ 2-2-5 = -\frac{1}{x+7}
 \\ \\ -5=-\frac{1}{x+7} \\ \\ 
-5(x+7)=-1 \\ \\ 
-5x-35=-1 \\ \\ 
-5x-35+1=0 \\ \\ 
-5x-34=0 \\ \\
-5x=34 \\ \\
x= \frac{34}{-5} \\ \\
x= -\frac{34}{5}$

$Espero$ $te$ $sea$ $de$ $utilidad,$ $mucho$ $gusto$ $y$ $hasta$ $pronto$

$"DIFUNDE$ $LA$ $CULTURA"$

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