a) (x^2+2x-15)/(2x^2-50) Zał: 2x^2 -50 ≠ 0 2x^2 ≠ 50 x^2 ≠ 25 x≠ -5 i x≠ 5 (x^2+2x-15)/(2x^2-50) = (x^2 +5x-3x-15) / [2(x-5)(x+5)] = [x(x+5) - 3(x+5)]/[2(x-5)(x+5)] = [(x+5)(x-3)] / [2(x-5)(x+5)] = (x-3)/[2(x-5)] = (x-3)/(2x-10) b) 2x^2-3x-9/4x^2-11x-3 Zał. 4x^2 - 11 x - 3 ≠ 0 Δ = (-11)^2 - 4*4*(-3) = 121 + 48 = 169 √Δ = 13 x1 = (11-13)/8 = -2/8 = -1/4 x2 = (11+13)/8 = 3 x≠ -1/4 i x ≠ 3 2x^2 -3x -9 = 0 Δ = (-3)^2 - 4*2*(-9) = 9 + 72 = 81 √Δ = 9 x1 = (3-9)/4 = 6/4 = 3/2 x2 = (3+9)/4 = 12/4 = 3 (2x^2-3x-9)/(4x^2-11x-3) = [2(x-3/2)(x-3)] / [4(x+1/4)(x-3)] = (2x-3)/(4x+1) c) x^3+3x^2-25x-75/x^2+8x+15 Zał x^2 +8x +15 ≠0 Δ = 8^2 -4*1*15 = 64 - 60 = 4 √Δ = 2 x1 = (-8-2)/2 = -5 x2 = (-8+2)/2 = -3 x≠ -5 i x ≠ - 3 (x^3+3x^2-25x-75)/(x^2+8x+15) = [(x^2(x+3) - 25(x+3)] / [(x+5)(x+3)] = [(x+3)(x^2-25)] / [(x+5)(x+3)] = [(x-5)(x+5)]/(x+5) = x-5 d) 25x^3+50x^2-x-2/5x^2+9x-2 Zał 5x^2 +9x - 2 ≠ 0 Δ = 9^2 - 4*5*(-2) = 81 + 40 = 121 √Δ = 11 x1 = (-9-11)/10 = -20/10 = -2 x2 = (-9+11)/10 = 2/10 = 1/5 x ≠ -2 i x ≠ 1/5 (25x^3+50x^2-x-2)/(5x^2+9x-2) = [25x^2(x+2) -(x+2)] / [5(x+2)(x-1/5)] = [(x+2)(25x^2-1)] / [5(x+2)(x-1/5)] = [(5x-1)(5x+1)] / (5x-1) = 5x+1
