Odpowiedź:
a) ... = [ ( 2 x - 1)² - x]*[ ( 2 x - 1)² + x ] = [4x²- 4x - 1 - x]*[ 4 x² - 4 x + 1 + x ] =
= ( 4 x² - 5 x - 1)*( 4 x²-3 x + 1 ) =
= 4*(x - [tex]\frac{5 - \sqrt{41} }{8})*( x - \frac{5 + \sqrt{41} }{8}[/tex]) *( 4 x² - 3 x + 1)
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bo
Δ[tex]_1 = 25 - 4*4*(-1) = 25 + 16 = 41[/tex]
√Δ[tex]_1 = \sqrt{41}[/tex]
[tex]x_1 = \frac{5 - \sqrt{41} }{8}[/tex] [tex]x_2 =[/tex] [tex]\frac{5 + \sqrt{41} }{8}[/tex]
Δ[tex]_2 = 9 - 4*4*1 = 9 - 16 < 0[/tex] - brak miejsc zerowych
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b) ... = [ (x - 2)² - (2 x + 1)²]*[ (x - 2)² + ( 2 x + 1)²] =
= [ x²- 4 x + 4 - 4 x²-4 x - 1 ]*[ x² - 4 x + 4 + 4 x² + 4 x + 1 ] =
= [ -3 x² - 8 x + 3 ]* [ 5 x² + 5 ] = -3*( x - [tex]\frac{1}{3}[/tex] )*( x + 3)* 5*( x² + 1) =
= - 15*( x - [tex]\frac{1}{3}[/tex] )*( x + 3)*( x² + 1 )
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bo Δ = (-8)² - 4*(-3)*3 = 64 + 36 = 100 √Δ = 10
[tex]x_1 = \frac{8 - 10}{- 6} = \frac{1}{3}[/tex] [tex]x_2 = \frac{8 + 10}{- 6} = - 3[/tex]
Szczegółowe wyjaśnienie: