Question

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Answer 1

Răspuns:

Explicație pas cu pas:

P(x) = 6x⁶ - 2x³ + 3x⁵ - 2x + 8 - 3x² - 6x⁶ + 2x - 11 + 4x³

a)

P(x) = 6x⁶ - 6x⁶ + 3x⁵ - 2x³ + 4x³ - 3x² - 2x + 2x + 8 - 11

P(x) = 3x⁵ + 2x³ - 3x² - 3

e)

P(1) = 3·1⁵ + 2·1³ - 3·1² - 3 = 3 + 2 - 3 - 3 = - 1

f)

Grad P(x) = 5

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P(x) + x² - 5x + 7 = 3x² + 2x - 1

P(x)  = 3x² + 2x - 1 -  x² + 5x - 7

P(x)  = 2x² + 7x - 8

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P(x) = x³ -2x² + 5x - 6

metoda 1: Restul impartirii la (x - 2) este egal cu P(2)

P(2) = 2³ -2·2² + 5·2 - 6 = 8 - 8 + 10 - 6 = 4

metoda 2:

P(x) = x³ -2x² + 5x - 6 = x³ -2x² + 5x - 10 + 4 = x²·(x - 2) + 5·(x - 2) + 4 = (x - 2)·(x² + 5) + 4  ⇒ restul impartirii la (x - 2) este 4

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$F(x) = \frac{\Big{x^2 -4}}{\Big{x^2 + 4x + 4}}$

a)

$F(3) = \frac{\Big{3^2 -4}}{\Big{3^2 + 4\cdot3 + 4}} = \frac{\Big{9 -4}}{\Big{9 + 12 + 4}} = \frac{\big5}{\Big{25}} = \frac{\Big1}{\Big5}$

b)

$F(x) = \frac{\Big{x^2 -4}}{\Big{x^2 + 4x + 4}} = \frac{(\Big{x -2})(\Big{x + 2})}{(\Big{x + 2})\Big{^2}} = \frac{\Big{x -2}}{\Big{x + 2}}$

c)

$F(x)\cdot(x-3) = \frac{ \Big {x-2}}{\Big {x+2}}\cdot(x -3) = \frac{( \Big {x-2})(\Big {x - 3})}{\Big {x+2}} = \frac{ \Big {x^2-5x+6}}{\Big {x+2}}$

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$E(x) = \Big (\frac{\Big {3}}{\Big {x-2}} - \frac{\Big {2}}{\Big {x+2}} - \frac{\Big {10}}{\Big {x^2-4}} \Big)\ \cdot \ \frac{\Big {x^2-4x+4}}{\Big {x}}=$

$= \Big (\frac{\Big {3}}{\Big {x-2}} - \frac{\Big {2}}{\Big {x+2}} - \frac{\Big {10}}{(\Big {x-2})(\Big{x+2})} \Big)\ \cdot \frac{(\Big {x-2})\Big{^2}}{\Big {x}} =$

$= \frac{\Big 3(\Big{x+2}) -\Big 2(\Big{x-2}) -\Big{10}} {(\Big {x-2})(\Big{x+2})} \ \cdot \ \frac{(\Big {x-2})\Big{^2}}{\Big {x}} =$

$= \frac{\Big{ 3x + 6 -2x + 4 -10}} {\Big{x+2}} \ \cdot \ \frac{\Big {x-2}}{\Big {x}} =$

$= \frac{\Big x} {\Big{x+2}} \ \cdot \ \frac{\Big {x-2}}{\Big {x}} = \frac{\Big {x-2}}{\Big{x+2}}$