Wyznacz trójmian kwadratowy o pierwiastkach x1,x2 i podanym zbiorze wartosci Y. a)x1=-4 x2=3 y=<-2,∞) b)x1=-4 x2=0 Y=(-∞,¼> . Rozwiąż równanie x²=(1-x)(1+x). Rozłóż wielomian w na czynniki; w(x)=-12x⁵+12x³-3x.Rozłóż wielomian w na czynniki, grupujac jego wyrazy w(x)½x³-⅙x²-3x+1, w(x)=x⁵+x⁴-2x³-2x²+x+1. oblicz log⅕ 5√5
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z.1
a) x1 = -4 oraz x2 = 3 oraz Y = < -2; +∞ )
Mamy
p = [x1+ x2]/2 = [-4 + 3]/2 = -1/2
q = -2
Mamy zatem
y = a(x -x1)(x-x2) = a(x - p)² + q
czyli
a( x +4)( x -3) = a[x +1/2]² - 2
a[x² +x -12] = a[x² +x + 1/4) - 2
a x² + a x - 12 a = a x² = ax + (1/4)a -2
-12 a -(1/4)a = -2
-12,25 a = -2
a = -2/(-12,25) = 8/49
zatem
y = (8/49)(x +4)(x -3) lub y = (8/49)x² +(8/49)x - 96/49
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b)
x1 = -4 oraz x2 = 0 oraz Y = ( -∞ ; 1/4>
Mamy
p = [x1 + x2]/2 = -4/2 = -2
q = 1/4
zatem
a(x-0)(x +4) = a[x +2]² + 1/4
ax(x +4) = a[x² +4x = 4] + 1/4
ax² + 4a x = a x² + 4a x + 4a + 1/4
0 = 4a + 1/4
4a = - 1/4 / : 4
a = - 1/16
mamy więc
y = (-1/16)x(x +4) lub y = (-1/16) x² - (1/4) x
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z.2
x² = (1 -x)( 1 +x)
x² = 1 - x²
2x² = 1
x² = 1/2
x = - 1/√2 lub x = 1/√2 lub inaczej x = - √2/2 lub x = √2/2
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W(x) = - 12 x⁵ + 12 x³ -3 x = -3x (4 x⁴ - 4x² + 1) = -3x ( 2x² - 1)² =
= -3x[ (√2 x -1)(√2 x + 1)]² = -3x (√3 -1)² * (√3 x + 1)²
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W(x) = x⁵ + x⁴ -2 x³ - 2 x² + x + 1 = x(x⁴ - 2x² + 1) + (x⁴ -2 x² + 1 ) =
= (x +1) (x² -1 )² = (x+1) (x-1)²(x+1)² = (x +1)³(x -1)²
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W(x) = (1/2) x³ - (1/6) x² - 3x +1 = (1/2) x[x² -6] -(1/6)[x² - 6] =
= (0,5x - 1/6) (x² - 6) = (0,5 x - 1/6)( x -√6)(x + √6)
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log ¹/₅ 5√5 = log ¹/₅ 5^(3/2) = (3/2) log ¹/₅ 5 = (3/2) *(-1) = (-3/2) = -1,5
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