Question

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Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\displaystyle a)~E(x)=\frac{(x+5)(2x-3)-2(x+2)^2+10}{x^2+2x} :\frac{x^3+12x^2-13x}{x^2+x-2} \\\\ E(x)=\frac{2x^2-3x+10x-15-2\Big(x^2+2 \cdot x \cdot 2+2^2\Big)+10}{x^2+2x} \cdot \frac{x^2+x-2}{x^3+12x^2-13x} \\\\ E(x)=\frac{2x^2+7x-15-2\Big(x^2+4x+4\Big)+10}{x(x+2)} \cdot \frac{(x-1)(x+2)}{x(x-1)(x+13)} \\\\ E(x)=\frac{2x^2+7x-15-2x^2-8x-8+10}{x} \cdot \frac{1}{x(x+13)} \\\\ E(x)=\frac{-x-13}{x\cdot x(x+13)} \Rightarrow E(x)=\frac{-(x+13)}{x^2(x+13)} \Rightarrow E(x)=-\frac{1}{x^2}$

$\displaystyle b)~-\frac{1}{E(x)} =16 \Rightarrow -\frac{1}{-\cfrac{1}{x^2} } =16 \Rightarrow 1 \cdot x^2=16 \Rightarrow x^2=16 \Rightarrow x=\pm \sqrt{16}\Rightarrow \\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=-4 \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=4$