z.1 (3 x⁴ - 27)/(x² - 3) DW = R { -√3 ; √3 } --------------------------- (3x⁴ - 27)/(x² -3) = [ 3(x² - 3)(x² +3)]/( x² - 3) = 3*(x² + 3) ======================================================== z.2 [x³ + 5x² - x - 5]/[x² + 4x - 5] = [ x(x² -1) +5(x² -1)]/[(x -1)(x +5)] = = [(x+5)(x² -1)]/ [(x-1)(x +5)] = [(x-1)(x +1)]/(x-1) = x + 1 DW = R {-5; 1} ======================================================== z.3 Czy wielomiany są równe? W(x) = (x +3)³ - 7( x² + 4) - 25 x = x³ +9x² +27x + 27 - 7x² -28 - 25x = = x³ + 2 x² + 2x -1 H(x) = (x+2)(x² + 2) - 5 = x³ +2x +2x² + 4 - 5 = x³ + 2x² + 2x - 1 W(x) = H ( x) =========================================================== z.4 W(x) = - 8x + 4 P(x) = x² + 2x - 1 Q(x) = 5x³ - x = 4 W(x)*P(x) = -8x³ -12 x² + 16 x - 4 zatem (1/4)*W(x)*P(x) = -2 x³ - 3 x² + 4x - 1 i ostatecznie (1/4)W(x)*P(x) - Q(x) = (-2 x³ - 3 x² + 4x -1) -(5x³ - x +4) = = - 7 x³ - 3 x² + 5x - 5 ====================================================== z.5 W(x) = -4 x⁴ + 26 x³ - 12 x² = -2x²(2 x² - 13 x +6) = = - 2 x² *2( x -6)(x - 1/2) = -2 x² (x -6)(2x - 1) ===================================================== z.6 Q(x) = x⁴ +(a +3)x³ + b x² + 16 P(x) = x⁴ - 8x² + 16 Aby wielomiany były równe muszą miec równe współczynniki przy takich samych potęgach, czyli a + 3 = 0 b = - 8 zatem a = -3 oraz b = - 8 Odp. b) ============================================================= z.7 W(x) = 8 x³ - 12 x² - 20 x + 30 = 2x(4x² - 10) -3(4 x² - 10) = = (2x - 3)(4x² - 10) = (2x -3)(2x - √10)(2x + √10) ============================================================== z.8 f(x) = 5x /[5x³ + 2 x² - 15 x - 6 ] Mianownik musi byc różny od 0. 5x³ + 2 x² - 15 x - 6 = 5x(x² -3) +2(x² - 3) = (5x +2)(x² - 3) = = (5x +2)(x -√3)(x + √3) 5x + 2 = 0 < => x = -2/5 x - √3 = 0 <=> x = √3 x = √3 = 0 <=. x = - √3 zatem dla x = -2/5 lub x = -√3 lub x = √3 mianownik byłby równy 0. Odp. D = R { -√3; - 2/5 ; √3} ================================== z.9 f(x) = √( -x³ - 4x + 2x² + 8 ) Wyrażenie pod pierwiastkiem musi być ≥ 0 -x³ - 4x + 2x² + 8 = -x ( x² = 4) +2( x² + 4) = (2 - x)(x² + 4) ≥ 0 <=> <=> 2 - x ≥ 0 <=> -x ≥ -2 <=> x ≤ 2 , bo x² + 4 > 0 dla każdej liczby x ∈ R. D = ( - ∞ ; 2 > ============================================================ z.10 W(x) = 2x³ + ax² - 6x -1 jest miejscem zerowym W , dlatego W(-1) = 0 Mamy więc 2*(-1)³ + a*(-1)² - 6*(-1) = 0 -2 +a + 6 = 0 a = -6 +2 = - 4 ==================== z.11 G(x) = (x² -4)(x² +5x + 6) Δ =25 - 4*1*6 = 25 - 24 = 1 x = [-5 -1]/2 = -3 lub x = [ -5 +1]/2 = -2 zatem G(x) =(x -2)(x+2)(x +3)(x +2) = (x +2)² (x -2)(x +3) ========================================================= z.12 W(x) = x² - 5x⁴ = x² ( 1 - 5 x²) = x² (1 - √5 x)(1 + √5 x) ======================================================== z.13 W(x) = 2 x⁵ - 2x + 2 V(x) = 2 x² + 3x - 3 Stopień wielomianu W(x) *V(x) = 7, bo 3 x⁵ * 2 x² = 6 x⁷ ================================================================ Zadania z 5 załącznika z.1 I x - 1 I ≥ 3 Odp. A ( - ∞ ; - 2 > u < 4 ; + ∞ ) --------------------------------------- z.2 L = 3a = 3 --> a = 1 P = [a²√3]/4 = [1²√3]/4 = √3/4 Odp. B ------------------------------------------- z.3 (x -2)² + (y + 3)² = 9 Punkt leży na osi OX ,gdy y = 0 zatem ( x - 2)² + ( 0 +3)² = 9 (x -2)² + 9 = 9 (x -2)² = 0 x = 2 zatem P = (2; 0) Odp.B ---------------------------------------- z.4 a1 = -1 oraz a2 = 2 zatem r = a2 - a1 = 2 - (-1) = 2 + 1 = 3 czyli a6 = a1 + 5r = -1 +5*3 = -1 + 15 = 14 Odp. C ------------------------------------------------- z.5 A = ( -2; 3) , prosta o równaniu y = (2/3) x + 5 (2/3)*a1 = -1 --> a1 = (-3/2) Prosta prostopadła do danej y = ( -3/2) x + b oraz A = ( -2; 3) zatem 3 = (-3/2)*(-2) + b 3 = 3 + b b = 0 y = (-3/2) x Odp.D ------------------------------------------------------------ z.6 4000 - 0,1 *4000 = 4000 - 400 = 3600 3600 - 0,1 *3600 = 3600 - 360 = 3 240 Odp. B ------------------------------------------------- z.7 [ x(x -2)]/3 + 4 = [(2x -3)(x - 1) -1 ]/6 + 2x / *6 2 x(x -2) + 24 = (2x - 3)(x - 1) - 1 + 12 x 2x² - 4x + 24 = 2x² - 2x - 3x + 3 - 1 + 12 x -4x + 24 = 7x + 2 7x + 4x = 24 - 2 11x = 22 / : 2 x = 2 ----- Odp.D ================================================================= cdn. za chwilę
