Do dziedziny wyrażenia wymiernego nie należą liczby, dla których mianownik ma wartość zero (ponieważ nie możemy dzielić przez zero). a) [latex]frac{3x^2+5}{2x-1} [/latex] [latex]2x-1 = 0[/latex] [latex]2x=1 /:2[/latex] [latex]x = frac{1}{2}[/latex] [latex]D = mathbb{R} ackslash {frac{1}{2}}[/latex] [latex]frac{3x^2+5}{2x-1} =_{dla x = 3} = frac{3 cdot 3^2+5}{2 cdot 3-1} = frac{3 cdot 9+5}{6-1} = frac{27+5}{5} = frac{32}{5} = 6,4[/latex] b) [latex]frac{x^2 + x - 1}{2x^2 + 5x} [/latex] [latex]2x^2 + 5x = 0[/latex] [latex]x cdot (2x + 5) = 0[/latex] [latex]x = 0 vee 2x + 5 = 0[/latex] [latex]x = 0[/latex] [latex]lub[/latex] [latex]2x + 5 = 0[/latex] [latex]2x = -5 /:2[/latex] [latex]x = - 2,5[/latex] [latex]D = mathbb{R} ackslash {-2,5; 0 } [/latex] [latex]frac{x^2 + x - 1}{2x^2 + 5x} =_{dla x = -2} = frac{(-2)^2 + (-2) - 1}{2cdot (-2)^2 + 5 cdot (-2)} = frac{4-2 - 1}{2cdot 4 -10} = frac{1}{8 -10} = frac{1}{-2} = -frac{1}{2} [/latex] c) [latex]frac{x+5}{x^3 -16x }[/latex] [latex]x^3 -16x =0[/latex] [latex]x cdot (x^2 - 16) =0[/latex] [latex]x cdot (x - 4) cdot (x + 4) =0[/latex] [latex]x = 0 vee x - 4 = 0 vee x + 4 = 0[/latex] [latex]x = 0 [/latex] [latex]lub[/latex] [latex]x - 4 = 0[/latex] [latex]x = 4[/latex] [latex]lub[/latex] [latex]x + 4 = 0 [/latex] [latex]x = - 4[/latex] [latex]D = mathbb{R} ackslash {-4; 0; 4 } [/latex] [latex]frac{x+5}{x^3 -16x } =_{dla x = 1} = frac{1+5}{1^3 -16 cdot 1} = frac{6}{1 -16}= frac{6}{-15} = -frac{2}{5} = -0,4[/latex] d) [latex]frac{(x^7-1)(x-2)}{x^2 + 3x + 2}[/latex] [latex]x^2 + 3x + 2 = 0[/latex] [latex]Delta = 9 - 8 = 1; sqrt{Delta} = 1[/latex] [latex]x_1 = frac{-3-1}{2} = frac{-4}{2} = -2[/latex] [latex]x_2= frac{-3+1}{2} = frac{-2}{2} = -1[/latex] [latex]D = mathbb{R} ackslash {-2; -1 } [/latex] Wartość wyrażenia: [latex]frac{(x^7-1)(x-2)}{x^2 + 3x + 2} [/latex] dla x = - 1 nie jest określona, bo - 1 ∉ D
a) (3x^2 + 5)/(2x-1) 2x-1 = 0 2x = 1 x = 1/2 D = R {1/2} dla x = 3 (3 *3^2 +5)/(2 *3 -1) = (27+5)/(6-1) = 32/5 = 6,4 b) (x^2 + x-1)/(2x^2 + 5x) 2x^2 + 5x = 0 x(2x+5) = 0 x = 0 lub 2x+5 = 0 2x = -5 x = -5/2 = -2,5 D = R {-2,5; 0} dla x = -2 [(-2)^2 +(-2) -1]/[2*(-2)^2 + 5*(-2)] = (4-2-1)/(2*4-10) = 1/(-2) = -1/2 c) (x+5)/(x^3 -16x) x^3 - 16x = 0 x(x^2 -16) = 0 x(x+4)(x-4) = 0 x = 0 v x = -4, v x = 4 D = R {-4; 0; 4} dla x = 1 (1+5)/(1^3 -16*1) = 6/(1-16) = 6/(-15) = -2/5 = - 0,4 d) [(x^7 -1)(x-2)]/(x^2 +3x+2) x^2 +3x+2 = 0 D(delta) = 3^2 -4*2 = 9-8 = 1 VD = 1 x1 = (-3-1)/2 = -2 x2 = (-3+1)/2 = -1 D = R {-2; -1} dla x = -1 nie jest określone,bo -1 nie należy do dziedziny (D)
