Zad 2) a) 5 3/5 : (-6/7) = 28/5 * (-7/6) = - -98/15 b) = 3/2 + 10/3*(-4/5) = 3/2 - 8/3 = 9/6 - 16/6 = -7/6 c) [latex]= sqrt[3]{27/8} - sqrt{25/4} =3/2-5/2=-2/2=-1[/latex] d) [latex] = (66:33)^{10} = 2^{10}=1024[/latex] Zad 3) a) .....⇔ 1 - 3x ≥ 4(x + 3) ⇔ 1 - 3x ≥ 4x + 12 ⇔ 7x ≤ -11 ⇔ x ≤ -11/7 b) .......... ⇔ x² + 12x + 36 > x² - 2 ⇔ 12x > -38 ⇔ x > - 38/12 ⇔ x > -19/6 Zad 1) a) = 16x² + 24x + 9 + 4x - x^2 = 15x² + 28x + 9 b) = 4x² - 12x + 9 - x - 10 = 4x² - 13x - 1 c) = x² - 1 + 2 = x² + 1
[latex]2.\a) 5frac{3}{5}:(-frac{6}{7}) = frac{28}{5}cdot(-frac{7}{6}) = -frac{196}{30} = -frac{98}{15} = -6frac{8}{15}\\b) 1frac{1}{2}+3frac{1}{3}:(-1frac{1}{4})=frac{3}{2}+frac{10}{3}:(-frac{5}{4})=frac{3}{2}+frac{10}{3}cdot(-frac{4}{5})=frac{9}{6}-frac{8}{3} = frac{9}{6}-frac{16}{6}=-frac{7}{6} = -1frac{1}{6}[/latex] [latex]c) sqrt[3]{3frac{3}{8}}-sqrt{6frac{1}{4}}=sqrt[3]{frac{27}{8}}-sqrt{frac{25}{4}} = frac{3}{2}-frac{5}{2}=-frac{2}{2} = -1[/latex] [latex]d) 66^{10}:33^{10} = (66:33)^{10} = 2^{10} = 1024[/latex] [latex]3.\a)\frac{1-3x}{4} geq x+3 |cdot4\\1-3x geq 4x + 12\\-3x-4x geq 12-1\\-7x geq 11 /:(-7)\\x leq -1frac{4}{7}[/latex] <------------------ //////////////////// | -------------------₀-----------₀--------> x -1⁴/₇ 0 kółeczko zaciemnione [latex]b\(x+6)^{2} extgreater x^{2}-2\\x^{2}+12x+36 extgreater x^{2}-2\\12x extgreater -2-36\\12x extgreater -38 /:12\\x extgreater -3frac{1}{6}[/latex] -------------------------------> |//////////////////////////////////// ---------------₀----------------₀-----------------> x -3¹/₆ 0 [latex]5.\a) (4x+3)^{2} + 4x-x^{2}=16x^{2}+24x+9 + 4x-x^{2}=15x^{2}+28x+9\\b) (2x-3)^{2}-(x+10) = 4x^{2}-12x+9 - x - 10 = 4x^{2}-13x-1\\c) (x-1)(x+1)+2 = x^{2}-1+2 = x^{2}+1[/latex]
