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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[tex]F(x) = \frac{\Big{x^2 -4}}{\Big{x^2 + 4x + 4}}[/tex]
a)
[tex]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}[/tex]
b)
[tex]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}}[/tex]
c)
[tex]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}}[/tex]
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[tex]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}}=[/tex]
[tex]= \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}} =[/tex]
[tex]= \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}} =[/tex]
[tex]= \frac{\Big{ 3x + 6 -2x + 4 -10}} {\Big{x+2}} \ \cdot \ \frac{\Big {x-2}}{\Big {x}} =[/tex]
[tex]= \frac{\Big x} {\Big{x+2}} \ \cdot \ \frac{\Big {x-2}}{\Big {x}} = \frac{\Big {x-2}}{\Big{x+2}}[/tex]
