Bonjour, On a P(x) = (x-1)(x²-ax+1) et Q(x) = (x-2)(x²-bx+c) 1) P(-1) = 2 ⇒ -2(1+a+1) = 2 ⇒ a = -3 ⇒P(x) = (x-1)(x²+3x+1) Q(0) = 1 ⇒ -2c = 1 ⇒ c = -1/2 Q(1)= 2 ⇒ -1(1-b+c)=2 ⇒ b-c = 3 ⇒ b = 5/2 on obtient P(x) = (x-1)(x²+3x+1) et Q(x) = (x-2)(x²-(5/2)x-(1/2)) 2) P(x) = Q(x) (x-1)(x²-ax+1) = (x-2)(x²-bx+c) x³- (1+a)x²+(a+1)x - 1 = x³ - (b+2)x² + (2b+c)x - 2c donc 1 = 2c ⇒ c = 1/2 et 1 + a = b+2 et 1 + a = 2b + c on peut en déduire que b + 2 = 2b + c ⇒ b = 2 + c = 2 + 1/2 ⇒ b = 3/2 et comme 1 +a = b + 2 ⇒ a = 5/2 on a alors P(x) = (x-1)(x² - (5/2)x + 1) et Q(x) = (x - 2)( x² - (3/2)x + 1/2 ) 4) P(-x) = -Q(x) (-x-1)(x²+ax+1) = (-x+2)(x²-bx+c) -x³ + (-a-1)x² + (-a-1)x - 1 = x³ + (b+2)x²-(c+2b)x-2c alors -2c = -1 ⇒ c = 1/2 -a 1 = b + 2 -a-1 = 2b +c on en déduit que b = 3/2 et a = -9/2 on a alors P(x) = (x-1)(x²+(3/2)x+1) et Q(x) = (x-2)(x²-(3/2)x+1/2 ) Bonnes fêtes
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Bonjour,On a P(x) = (x-1)(x²-ax+1) et Q(x) = (x-2)(x²-bx+c)
1)
P(-1) = 2 ⇒ -2(1+a+1) = 2 ⇒ a = -3 ⇒ P(x) = (x-1)(x²+3x+1)
Q(0) = 1 ⇒ -2c = 1 ⇒ c = -1/2
Q(1)= 2 ⇒ -1(1-b+c)=2 ⇒ b-c = 3 ⇒ b = 5/2
on obtient
P(x) = (x-1)(x²+3x+1) et Q(x) = (x-2)(x²-(5/2)x-(1/2))
2)
P(x) = Q(x)
(x-1)(x²-ax+1) = (x-2)(x²-bx+c)
x³- (1+a)x²+(a+1)x - 1 = x³ - (b+2)x² + (2b+c)x - 2c
donc
1 = 2c ⇒ c = 1/2
et
1 + a = b+2 et 1 + a = 2b + c
on peut en déduire que
b + 2 = 2b + c ⇒ b = 2 + c = 2 + 1/2 ⇒ b = 3/2
et comme
1 +a = b + 2 ⇒ a = 5/2
on a alors
P(x) = (x-1)(x² - (5/2)x + 1) et Q(x) = (x - 2)( x² - (3/2)x + 1/2 )
4)
P(-x) = -Q(x)
(-x-1)(x²+ax+1) = (-x+2)(x²-bx+c)
-x³ + (-a-1)x² + (-a-1)x - 1 = x³ + (b+2)x²-(c+2b)x-2c
alors
-2c = -1 ⇒ c = 1/2
-a 1 = b + 2
-a-1 = 2b +c on en déduit que b = 3/2 et a = -9/2
on a alors
P(x) = (x-1)(x²+(3/2)x+1) et Q(x) = (x-2)(x²-(3/2)x+1/2 )
Bonnes fêtes