Question

E(x)=
$( \frac{ - x { }^{2 } -3x + 4 }{x {}^{2} - 16 } - \frac{1}{x - 4}) \div \frac{x}{3(x - 4)}$
aratati ca E(x)=-3
pentru orice x€R\{-4, 0, 4}
Dau coroana

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Answer 1

Răspuns:

Explicație pas cu pas:

$(\frac{-x^{2}-3x+4 }{x^{2} -16 } -\frac{1}{x-4} ):\frac{x}{3(x-4)} =\\ (\frac{-x^{2}-3x+4 }{x^{2} -4^{2}  }-\frac{1}{x-4} ):\frac{x}{3(x-4)}=\\  (\frac{-x^{2}-3x+4 }{(x+4)(x-4)  }-\frac{1}{x-4} ):\frac{x}{3(x-4)}=\\ (\frac{-x^{2}-3x+4 }{(x+4)(x-4)  }-\frac{x+4}{(x+4)(x-4)} ):\frac{x}{3(x-4)}=\\ (\frac{-x^{2}-3x+4-x-4 }{(x+4)(x-4)} ):\frac{x}{3(x-4)} =\\ \frac{-x^{2} -4x}{(x+4)(x-4)} * \frac{3(x-4)}{x} =\\ \frac{-x(x+4)}{(x+4)(x-4)} * \frac{3(x-4)}{x} = -3$

Answer 2

Răspuns:

Explicație pas cu pas:

E(x) = [ ( - x² - 3 x + 4)/ (x²-16) -  1 /(x - 4) ] : x /(x-4)

E(x) = [ (-x²-3x + 4)/(x-4)(x+4) - ( x+4)/(x-4)(x+4) ] × 3(x-4)/x

E(x) = [ ( - x²- 3x + 4 - x - 4) /(x-4)(x+4) ] × [3(x-4)/x]

E(x) = [ ( - x² - 4 x ) /(x+4) ] × (3/x)

E(x) = - 3 x(x+4)/x(x+4)

E(x) = - 3   ∀  x ∈ R\ { -4, 0, 4}