1. 2x³-7x²-17x+10 = 0 (x-5)(2x²+3x-2) = 0 x-5 = 0 => x = 5 lub 2x²+3x-2 = 0 Δ = 9+16 = 25 √Δ = 5 x1 = (-3-5)/4 = -8/4 = -2 x2 = (-3+5)/4 = 2/4 = 1/2 x = -2 v x = 1/2 v x = 5 2. x³-4x²+8 = 0 (x-2)(x²-2x-4) = 0 x-2 = 0 => x = 2 lub x²-2x-4 = 0 Δ = 4+16 = 20 √Δ = 2√5 x1 = (2-2√5)/2 = 1-√5 x2 = (2+2√5)/2 = 1+√5 x = 1-√5 v x = 1+√5 v x = 2 3. x³-13x-12 > 0 (x+3)(x²-3x-4) > 0 x+3 = 0 => x = -3 lub x²-3x-4 = 0 Δ = 9+16 = 25 √Δ = 5 x1 = (3-5)/2 = -2/2 = -1 x2 = (3+5)/2 = 8/2 = 4 x ∈ (-3; -1) v (4; +∞) 4. (x+1)/x > 0 Z: x ≠ 0 x(x+1) = 0 x = 0 lub x+1 = 0 => x = -1 x ∈ (-∞; -1) v (0; +∞) Lub x+1/x > 0 (x²+1)/x > 0 x(x²+1) > 0 x²+1 - zawsze > 0 x > 0 ====
1. (2x³-7x²-17x+10):(x-5)=2x²+3x-2 -2x³+10x² --------------- 3x²-17x -3x²+15x ----------- -2x+10 2x-10 --------- = = (2x²+3x-2)(x-5)=0 Δ=9+4*2*2=25 x=(-3-5):4=-2 v x=(-3+5):4=1/2 v x=5 2. (x³-4x²+8):(x-2)=x²-2x-4 -x³+2x² ----------- -2x²+8 2x²+4x ----------- -4x+8 4x-8 ________ = = (x²-2x-4)(x-2)=0 Δ=4+4*4=20 √Δ=√20=2√5 x=(2-2√5):2=1-√5 v x=1+√5 v x=2 3. x³-13x-12>0 x³-x-12x-12>0 x(x²-1)-12(x+1)>0 x(x+1)(x-1)-12(x+1)>0 (x+1)[x(x-1)-12]>0 (x+1)(x²-x-12)>0 Δ=1+4*12=49 m.z. x(1-7):2=-3 v x=(1+7):2=4 v x=-1 (x+3)(x+1)(x-4)>0 badamy znaki w przedzialach (rys. w zalaczniku) Odp. x∈(-3,-1)u(4,+∞) 4. x+1/x>0/*x² zal. x≠0 x³+x>0 x(x²+1)>0 x>0 , bo (x²+1 zawsze >0)
