PROSZE O SZYBKIE ROZWIAZANIE! BARDZO PILNE! DZIEKUJE! ZADANIA W ZALĄCZNIKU: nr.29(b, d, f, h) nr.30 (b, d, f, h)

29. b) [latex]x^5 - 33 = 0[/latex] [latex]x^5 = 33[/latex] d) [latex]3x^2- frac{1}{9} = 0[/latex] [latex]3x^2= frac{1}{9} [/latex] [latex] x^{2} = frac{1}{27} [/latex] f) [latex]8x^7- frac{1}{16} = 0[/latex] [latex]8x^7= frac{1}{16} [/latex] [latex]x^7= frac{1}{128} [/latex] h) [latex]-4x^5+9=1[/latex] [latex]-4x^5=-8[/latex] [latex]x^5=2[/latex] 30. b) [latex]-x^6+16=0[/latex] [latex]-x^6=-16[/latex] [latex]x^6=16[/latex] d) [latex]2x^4-32=0[/latex] [latex]2x^4=32[/latex] [latex]x^4=16[/latex] f) [latex]3x^8+ frac{1}{4} =0[/latex] [latex]3x^8=- frac{1}{4} [/latex] [latex]x^8=- frac{1}{12} [/latex] h) [latex] frac{1}{9}x^6+3=6 [/latex] [latex] frac{1}{9} x^6=3[/latex] [latex]x^6=27[/latex]

29 [latex]b)\x^5-33=0\\x^5=33\\x= sqrt[5]{33} \\d)\3x^3- frac{1}{9}=0 \\3x^3= frac{1}{9} /:3\\x^3= frac{1}{27}\\x= frac{1}{3} [/latex] [latex]f)\8x^7- frac{1}{16} =0\\8x^7= frac{1}{16} /:8\\x^7= frac{1}{16cdot 8} \\x^7= frac{1}{2 cdot 2 cdot 2cdot 2cdot 2cdot 2cdot 2}\\ x^7= frac{1}{2^7} \\x=frac{1}{7}[/latex] [latex]h)\-4x^5+9=1\ \ -4x^5=1-9\\-4x^5=-8 /:(-4)\\x^5=2\\x= sqrt[5]{2} [/latex] 30 [latex]b)\-x^6+16=0\\-x^6=-16 /:(-1)\ \x^6=16\\x_1=- sqrt[6]{16}=- sqrt[6]{4^2}=- sqrt[3]{4},qquad x_2= sqrt[6]{16}= sqrt[6]{4^2}= sqrt[3]{4}[/latex] [latex]d)\2x^4-32=0\\2x^4=32 /:2\\x^4=16\ \x_1=- sqrt[4]{16}=-2, qquad x_2= sqrt[4]{16}=2[/latex] [latex]f)\3x^8+ frac{1}{4} =0\\3x^8=- frac{1}{4} /:3\\x^8 eq - frac{1}{12}[/latex] równanie sprzeczne, bo nie istnieje parzysty pierwiastek z liczby ujemnej [latex]h)\ frac{1}{9} x^6+3=6\\ frac{1}{9} x^6=6-3\\ frac{1}{9} x^6=3 /*9\\ x^6=27\\x_1= sqrt[6]{27}= sqrt[6]{3^3}= sqrt[]{3}, qquad x_2=- sqrt[6]{27} =- sqrt[6]{3^3}=- sqrt{3}[/latex]