Bardzo pilne prosze pomóżcie!. 8.8    wyznacz sumę, iloczyn i obie różnice przedziałów A i B gdy A=(-3:4), B=<1;7> 5.26 rozwiąż równanie : a) (2x-1)(3-x)(x+4)=0 5.28 rozwiąż równanie: d) (x-3)do potęgi 3 =243 h) x do potęgi 3 - x do potęgi 2=0

Zad. 8.8 [latex]A = (-3; 4), B=langle1; 7 angle[/latex]   [latex]A cup B = (-3; 4) cup langle1; 7 angle = (-3; 7 angle[/latex] [latex]A cap B = (-3; 4) cap langle1; 7 angle = langle 1; 4)[/latex] [latex]A ackslash B = (-3; 4) ackslash langle1; 7 angle = (-3; 1)[/latex] [latex]B ackslash A = langle1; 7 angle ackslash (-3; 4) = langle 4; 7 angle[/latex]   Zad. 5.26 a) [latex](2x-1)(3-x)(x+4)=0[/latex] [latex]2x-1 = 0 vee 3-x = 0 vee x+4=0[/latex]   [latex]2x-1 = 0[/latex] [latex]2x=1 / : 2[/latex] [latex]x = frac{1}{2}[/latex]   [latex]3-x = 0[/latex] [latex]-x = - 3 / cdot (-1)[/latex] [latex]x = 3[/latex]   [latex]x+4=0[/latex] [latex]x = - 4[/latex]   Odp. [latex]x = - 4 lub x = frac{1}{2} lub x = 3[/latex]   Zad. 5.28 d) [latex](x-3)^3 = 243[/latex] [latex](x-3)^3 - 243 = 0[/latex] [latex](x-3)^3 - (sqrt[3]{243})^3 = 0[/latex] [latex][(x-3) - sqrt[3]{243}][(x-3)^2 + (x-3) cdot sqrt[3]{243} + (sqrt[3]{243})^2] = 0[/latex] [latex](x-3 - sqrt[3]{243})(x^2-6x + 9 + sqrt[3]{243}x-3sqrt[3]{243} + sqrt[3]{243^2}) = 0[/latex] [latex](x-3 - sqrt[3]{3^5})(x^2-6x + sqrt[3]{3^5}x+ 9-3sqrt[3]{3^5} + sqrt[3]{(3^5)^2}) = 0[/latex] [latex](x-3 - 3^{frac{5}{3}})(x^2-6x + 3^{frac{5}{3}}x+ 9-3cdot 3^{frac{5}{3}}+ sqrt[3]{3^{10}}) = 0[/latex] [latex](x-3 - 3^{frac{5}{3}})[x^2-(6 - 3^{frac{5}{3}})x+ 9-3^{frac{8}{3}}+ 3^{frac{10}{3}}] = 0 [/latex] [latex]x-3 - 3^{frac{5}{3}} =0 vee x^2-(6 - 3^{frac{5}{3}})x+ 9-3^{frac{8}{3}}+ 3^{frac{10}{3}}= 0 [/latex]   [latex]x-3 - 3^{frac{5}{3}} =0[/latex] [latex]x=3 + 3^{frac{5}{3}}[/latex]   [latex]x^2-(6 - 3^{frac{5}{3}})x+ 9-3^{frac{8}{3}}+ 3^{frac{10}{3}}= 0[/latex] [latex]Delta = [-(6 - 3^{frac{5}{3}})]^2 - 4 cdot 1 cdot (9-3^{frac{8}{3}}+ 3^{frac{10}{3}}) = (3^{frac{5}{3}} - 6)^2-36+4 cdot 3^{frac{8}{3}} - 4 cdot 3^{frac{10}{3}} =[/latex] [latex]= 3^{frac{10}{3}} - 12 cdot 3^{frac{5}{3}} + 36 -36+4 cdot 3^{frac{8}{3}} - 4 cdot 3^{frac{10}{3}} = - 4 cdot 3 cdot 3^{frac{5}{3}} +4 cdot 3^{frac{8}{3}} - 3 cdot 3^{frac{10}{3}} = [/latex] [latex]= - 4 cdot 3^{frac{8}{3}} +4 cdot 3^{frac{8}{3}} - 3^{frac{13}{3}} =- 3^{frac{13}{3}} < 0[/latex] Równanie nie ma rozwiązań   Odp. [latex]x=3 + 3^{frac{5}{3}}[/latex]   h) [latex]x^3 - x ^2=0[/latex] [latex]x^2 cdot (x - 1)=0[/latex] [latex]x^2 = 0 vee x - 1=0[/latex]   [latex]x^2 = 0[/latex] [latex]x = 0[/latex]   [latex]x - 1=0[/latex] [latex]x = 1[/latex]   Odp. [latex]x = 0 lub x = 1[/latex]