Proszę o pomoc!!potrzebuje rozwiązania tych zadań na jutro gdzieś do godz.10 Zadania w załączniku
Proszę o pomoc!!potrzebuje rozwiązania tych zadań na jutro gdzieś do godz.10
Zadania w załączniku
Odpowiedź
1. a) f(x) = (2x+4)/(3x+12) 3x+12 = 0 3x = -12 /:3 x = -4 D= R {-4} b) g(x) = (x^2 + 5x - 4)/(x^3 - 2x^2 - 4x + 8) x^3 - 2x^2 - 4x + 8 = 0 x^2*(x-2) - 4(x-2) = 0 (x^2 - 4)(x-2) = 0 (x+2)(x-2)(x-2) = 0 x = -2 v x = 2 D = R {-2,2} 2. f(x) = (x^2 - 6x + 8)/(x^2 - 4) x^2 - 4 = 0 (x+2)(x-2) = 0 D = R {-2,2} x^2 - 6x + 8 = 0 D(delta) = b^2 - 4ac = 36-32 = 4, VD = 2 x1 = (-b-VD)/2a = (6-2)/2 = 2 x2 = (-b+VD)/2a = (6+2)/2 = 4 x1 = 2 nie należy do dziedziny,zatem: MZ: 4 3. F(x) = (2x+3)/(x+2), dla x = V5 - 1 Z: x =/= -2 F(V5-1) = [2(V5-1)+3]/(V5-1+2) = (2V5-2+3)/(V5+1) = (2V5+1)/(V5+1) *(V5-1)/(V5-1) = = (2V5+1)(V5-1)/4 = (10-2V5+V5-1)/4 = (9-V5)/4 4. MZ: 1, -2 D = R {-5, 0,3} np.: f(x) = (x-1)(x+2)/(x+5)(x-0,3) f(x) = (x^2 + x - 2)/(x^2 + 4,7x - 1,5) 5. a) (3x+2)/(x+1) = (x-2)/(x-1) = x+1 = 0 => x = -1 x-1 = 0 => x = 1 D = R {-1,1} = (x-1)(3x+2) = (x+1)(x-2) 3x^2 + 2x - 3x - 2 = x^2 - 2x + x - 2 2x^2 = 0 x = 0 b) 1 + 1/(x-2) = 3/(x-1) x-1 = 0 => x = 1 x-2 = 0 => x = 2 D = R {1,2} (x-2+1)/(x-2) = 3/(x-1) (x-1)/(x-2) = 3x/(x-2) (x-1)^2 = 3(x-2) x^2 - 2x +1 = 3x - 6 x^2 - 5x + 7 = 0 D = 25 -28 = -3 D < 0 a = 1 > 0 ramiona paraboli skierowane w gó x e (-oo,1) u (2,+oo) c) (x-5)(4+2x)/(5x^2 +1) = 0 5x^2 +1 > 0 D = R (x-5)(4+2x) = 0 x1 = 5 ===== lub 4+2x = 0 /:2 x = -2 ===== x = -2, v x = 5 6. a) (x+1)/2 * 10x/(x^2 + x) = x^2 + x = 0 x(x+1) = 0 x = 0 v x = -1 D = R {-1,0} = 5x(x+1)/x(x+1) = 5 b) x^5/(2x-6) : x^3/(x^2 - 6x + 9) = 2x-6 = 0 /:2 x = 3 x^2 - 6x + 9 = 0 D(delta) = 36-36 = 0 x = 6/2 = 3 D = R {3} = x^5/(2x-6) *(x^2 - 6x + 9)/x^3 =x^2/2(x-3) * (x-3)^2 = x^2(x-3)/2 = (x^3 - 3x^2)/2
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