Zad\u00e1n\u00ed:<\/strong><\/p>\n Zapi\u0161 ve tvaru mocniny s racion\u00e1ln\u00edm mocnitelem: $$\\sqrt[3]{{x}^{7}}$$<\/p>\n <\/p>\n \u0158e\u0161en\u00ed:<\/strong><\/p>\n $$\\sqrt[3]{{x}^{7}}={({x}^{7})}^{\\frac{1}{3}}={x}^{\\frac{7}{3}}$$<\/p>\n Zad\u00e1n\u00ed:<\/strong><\/p>\n \u010c\u00e1ste\u010dn\u011b odmocni: $$\\sqrt{300}$$<\/p>\n <\/p>\n \u0158e\u0161en\u00ed:<\/strong><\/p>\n $$\\sqrt{300}=\\sqrt{3\\cdot100}=\\sqrt{3\\cdot4\\cdot25}=\\sqrt{3\\cdot2\\cdot2\\cdot5\\cdot5}=2\\cdot5\\cdot\\sqrt{3}=10\\cdot\\sqrt{3}$$<\/p>\n Zad\u00e1n\u00ed: <\/strong>Pro \ud835\udc4e \u2208 N <\/strong>upravte v\u00fdraz a vyj\u00e1d\u0159ete jej ve tvaru odmocniny<\/strong> o z\u00e1kladu a:<\/strong><\/p>\n $${{a}^{\\frac{1}{4}}}:\\sqrt[6]{a}$$<\/p>\n <\/p>\n \u0158e\u0161en\u00ed:<\/strong><\/p>\n Odmocninu m\u016f\u017eeme vyj\u00e1d\u0159it pomocn\u00ed mocniny, v\u00fdraz tedy uprav\u00edme n\u00e1sledovn\u011b:<\/p>\n $${{a}^{\\frac{1}{4}}}:\\sqrt[6]{a}={{a}^{\\frac{1}{4}}}:{{a}^{\\frac{1}{6}}}$$<\/p>\n Podle pravidla: $$\\frac{{a}^{m}} {{a}^{n}}= {a}^{m-n}$$ p\u0159ep\u00ed\u0161eme na:<\/p>\n $${a}^{{\\frac{1}{4}}-{\\frac{1}{6}}}$$<\/p>\n n\u00e1sledn\u011b od sebe zlomky ode\u010dteme a dostaneme:<\/p>\n $${a}^{{\\frac{1}{4}}-{\\frac{1}{6}}}={a}^{\\frac{1}{12}}$$<\/p>\n Zb\u00fdv\u00e1 posledn\u00ed krok, kter\u00fdm je p\u0159eveden\u00ed mocniny na odmocninu.<\/p>\n Vych\u00e1z\u00edme z pravidla: $${a}^{\\frac{m}{n}}=\\sqrt[n]{{a}^{m}}$$<\/p>\n A p\u0159evedeme mocninu na odmocninu:<\/p>\n $${a}^{\\frac{1}{12}}=\\sqrt[12]{a}$$<\/p>\n V\u00fdsledkem je tedy: $$\\sqrt[12]{a}$$<\/p>\n","protected":false},"comment_status":"closed","ping_status":"closed","template":"","class_list":["post-1090","lesson","type-lesson","status-publish","hentry"],"yoast_head":"\nP\u0159\u00edklad 2<\/h2>\n
P\u0159\u00edklad 3<\/h2>\n