Task/25404599
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<span>Доказать , что функция f(x)=(x+4)|x-5|+(x-4)|x+5| является нечётной.
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* * * f(-5) = -10 ; f(5) =10 ; f(0) =4*5 - 4*5 = 0. * * *
a) x ≥ 5 .
f(x) = (x+4)*(x -5) + (x - 4)*(x +<span>5) </span>= 2(x² - 20) .
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b) x ≤ - 5 .
f(x) = (x+4)*(-(x-5)) + (x- 4)*(-(x+5) ) = - ( (x+4)*(x-5) +(x - 4)*(x+5) ) =
= - 2(x² -20) .
f(-x₁) = - f(x₁) , т.к. если x₁ ≤ - 5 ⇒ - x₁ ≥ 5 .
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c) - 5 < x < 5
f(x) = (x+4)*(- (x-5) ) + (x - 4)*(x +5) = - (x+4)*(x - 5) + (x - 4)*(x +5) =
= 2x .
Значит , если - 5 < x₀ ≤ 0 ,то 0 ≤ - x₀ < 5
f(- x₀) =-2x₀ = - 2f(x₀) .
функция <span> f(x)=(x+4)|x-5|+(x-4)|x+5| </span>является <span> нечётной.</span>
-2(x² -20) 2x 2x 2(x² -20)
///////////////////////// [-5] ||||||||||||||| [0] ||||||||||||||| [5] /////////////////////////
* * * * * * *P.S.* * * * * * *
f(-5) = -2((-5)² -20) =10 или f(-5) =2*(-5) = - 10 .
f(5) =2(5² -20) =10 или f(5) =2*5 =10.
f(0) =2*0 =2*(-0) =0 .
A1+a1+d+a1+2d=21⇒3a1+3d=21⇒a1+d=7⇒d=7-a1
{a1+d-1=a1q⇒a1+7-a1-1=a1q⇒a1q=6
a1+2d+1=a1q²⇒a1+14-2a1+1=a1q²⇒a1q²=15-a1
15-a1=6q
q=(15-a1)/6
a1(15-a1)/6=6
15a1-a1²=36
a1²-15a1+36=0
(a1)1+(a1)2=15 U (a1)1*(a1)2=36
(a1)1=3⇒d1=7-3=4
(a1)2=12⇒d2=7-12=-5
Ответ 3;7;11 или 12;7;2
<span>х</span>² <span>- 7х < 6х - 15 - х</span>²
x² + x² - 7x - 6x + 15 <0
2x² - 13x + 15 < 0
2x² - 13x + 15 = 0
D= 169 - 120 = 49
x₁ = 13 + 7 / 4 = 5
х₂= 13 - 7 / 4 = 1,5
2(х-5)(х-1,5) < 0
Ответ: (1,5;5).