Question

$E(x) = ( \frac{x}{x-4} - \frac{2}{x+4} + \frac{7}{16- x^{2}}) : \frac{ x^{2} + 2x +1}{ x^{2} - 8x + 16}$, x ∈ R \ {-4; -1; 4}. Aduceţi E(x) la forma cea mai simplă.

Answer 1

$E(x)=(\frac{x}{x-4}^{(x+4}-\frac{2}{x+4}^{(x-4}-\frac{7}{x^2-16}):\frac{(x+1)^2}{(x-4)^2}= \\\\=\frac{x(x+4)-2(x-4)-7}{x^2-16}*\frac{(x-4)^2}{(x+1)^2}= \\\\ =\frac{x^2+4x-2x-8-7}{\not{(x-4)}(x+4)}*\frac{(x-4)^\not2}{(x+1)^2}= \\\\= \frac{x^2+2x-15}{x+4}*\frac{x-4}{(x+1)^2} \\\\\\ x^2+2x-15 \\\\ a=1\\ b=2\\c=-15\\\Delta=2^2-4*(-15)=4+60\to\boxed{64} \\\\ x_1;x_2=\frac{-b\pm \sqrt{\Delta}}{2a}=\frac{-2\pm\sqrt{64}}{2} \\\\\\ x_1=\frac{-2+8}{2}=\frac{\not6}{\not2}\to 3$

$x_2=\frac{-2-8}{2}=\frac{-\not10}{\not2}\to-5 \\\\\\ x^2+2x-15=(x-3)(x+5) \\\\\\ E(x)=\frac{(x-3)(x+5)}{x+4}*\frac{x-4}{(x+1)^2} \\\\\\ \boxed{\boxed{E(x)=\frac{(x-3)(x+5)(x-4)}{(x+4)(x+1)^2}}}$

Answer 2

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