(9x²-4)(2x²-7x+3)=0 9x²-4 = 0 ∨ 2x²-7x+3=0 x² = ⁴/₉ x = ²/₃ ∨ x = -²/₃ 2x²-7x + 3 =0 Δ = 49-4*3*2 = 49-24 = 25 √Δ = 5 x₁ =(7-5)/4 = ¹/₂ x₂ = (7+5)/4 = 12/4 = 3 Jedynym całkowitym pierwiastkiem równania jest 3. 2. a) u(x)=x³ - 5x² + x - 5 = x²(x-5)+x-5 = (x-5)(x²+1) b) u(x)=3x³ - x² + x -3 =(x-1)(3x²+2x+3) Δ 4-4*3*3 < 0
[latex]1.\\(9x^{2}-4)(2x^{2}-7x+3) = 0\\2x^{2}-7x+3 = 0\\Delta = 49-24 = 25\\sqrt{Delta} = 5\\x_1 = frac{7-5}{4} = frac{1}{2} eq C\\x_2 = frac{7+5}{4} = frac{12}{4} = 3[/latex] [latex]2(3x+2)(3x-2)(x-frac{1}{2})(x-3) = 0\\x = -frac{2}{3} vee x = frac{2}{3} vee x = frac{1}{2} vee x = 3\\x = 3 in C[/latex] [latex]2.\a)\\U(x) = x^{3}-5x^{2}+x-5 = x^{2}(x-5) + (x-5) = (x^{2}+1)(x-5)[/latex] [latex]b)\\U(x) = 3x^{3}-x^{2}+x-3 =(x-1)(3x^{2}+2x+3)[/latex]
