Question

POR FAVOR ME AJUDEM..

c0dbbc080dc8709575a174b04ca833c8.docx

Answer 1

$a) \\ \\ \dfrac{3x^2 - 12}{6x + 12} \\ \\ \\ \dfrac{3(x - 2)(x \not + 2)}{6(x \not + 2)} \\ \\ \\ \dfrac{3(x - 2)}{6} \\ \\ \\ => \dfrac{x - 2}{2}$

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$b) \\ \\ \dfrac{x^4 - 1}{x^3 + x} \\ \\ \\ \dfrac{(x + 1)(x -1)(x^2 + 1)}{x(x^2 + 1)} \\ \\ \\ \dfrac{(x + 1)(x -1)(x^2\not + 1)}{x(x^2\not + 1)} \\ \\ \\ \dfrac{(x + 1)(x -1)}{x} \\ \\ \\ => \dfrac{x^2 - 1}{x}$

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$c) \\ \\ \\ \dfrac{x^2 + 2x + 1}{x + 1} \\ \\ \\ \dfrac{(x + 1)(x + 1)}{(x + 1)} \\ \\ \dfrac{(x + 1)(x \not+ 1)}{(x\not + 1)} \\ \\ \\ => x + 1$

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$d) \\ \\ \dfrac{15x^2 - 15y^2}{6x^2 + 12xy + 6y^2} \\ \\ \\ \dfrac{15(x^2 - y^2)}{6 (x^2 + 2xy + y^2)} \\ \\ \\ \dfrac{15(x - y)(x + y)}{6 (x + y)(x + y)} \\ \\ \\ \dfrac{15(x - y)(x \not+ y)}{6 (x + y)(x\not + y)} \\ \\ \\ \dfrac{15(x - y)}{6 (x + y)} \\ \\ \\=> \dfrac{5(x - y)}{2(x + y)}$