Beklager kodinga. Det ble litt for avansert for meg
Oppgaven lyder:
(2^3/4 * 2^4/3) / 4^3/2. Eller:
[tex] (2^(\frac{3}{4}) * 2^(\frac{4}{3}) )/ {4^(\frac{3}{2}) [/tex]
Hvordan går man fram?
Regning med røtter. Litt hjelp til en oppgave.
Moderators: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
Stemmer det. Og nevneren blir det samme som:
[tex]4^{\frac{3}{2}}=\sqrt{4^3}=\sqrt{64}=8[/tex]
husk at:
[tex]2^{\frac{25}{12}}=2*2*2^{\frac{1}{12}}\\ fordi\; 2*2*2^{\frac{1}{12}}=2^{\frac{12}{12}+\frac{12}{12}+\frac{1}{12}}=2^{\frac{25}{12}}[/tex]
[tex]8=2*2*2[/tex]
Regnestykket blir til slutt:
[tex]\frac{2*2*2^{\frac{1}{12}}}{2*2*2}[/tex]
Så kan du regne ut resten.
[tex]4^{\frac{3}{2}}=\sqrt{4^3}=\sqrt{64}=8[/tex]
husk at:
[tex]2^{\frac{25}{12}}=2*2*2^{\frac{1}{12}}\\ fordi\; 2*2*2^{\frac{1}{12}}=2^{\frac{12}{12}+\frac{12}{12}+\frac{1}{12}}=2^{\frac{25}{12}}[/tex]
[tex]8=2*2*2[/tex]
Regnestykket blir til slutt:
[tex]\frac{2*2*2^{\frac{1}{12}}}{2*2*2}[/tex]
Så kan du regne ut resten.
Takk!alexleta wrote:Stemmer det. Og nevneren blir det samme som:
[tex]4^{\frac{3}{2}}=\sqrt{4^3}=\sqrt{64}=8[/tex]
husk at:
[tex]2^{\frac{25}{12}}=2*2*2^{\frac{1}{12}}\\ fordi\; 2*2*2^{\frac{1}{12}}=2^{\frac{12}{12}+\frac{12}{12}+\frac{1}{12}}=2^{\frac{25}{12}}[/tex]
[tex]8=2*2*2[/tex]
Regnestykket blir til slutt:
[tex]\frac{2*2*2^{\frac{1}{12}}}{2*2*2}[/tex]
Så kan du regne ut resten.
[tex]\frac{2^{\frac{1}{12}}}{2^{\frac{12}{12}}}[/tex]
1-12=-11. [tex]2^{-\frac{11}{12}}[/tex] ? Kan jeg spørre om å få forklart den siste biten?
La oss analysere hele:
[tex]\frac{2^{\frac{3}{4}}*2^{\frac{4}{3}}}{4^{\frac{3}{2}}}[/tex]
Teller:
[tex]2^{\frac{3}{4}}*2^{\frac{4}{3}}=\sqrt[4]{2^{3}}*\sqrt[3]{2^{4}}=2^{\frac{3}{4}+\frac{4}{3}}=2^{\frac{9}{12}+\frac{16}{12}}=2^{\frac{25}{12}}=2^{\frac{12}{12}}*2^{\frac{12}{12}}*2^{\frac{1}{12}}=2*2*2^{\frac{1}{12}}[/tex]
Nevner:
[tex]4^{\frac{3}{2}}=\sqrt{4^{3}}=\sqrt{64}=8=2*2*2[/tex]
Brøk:
[tex]\frac{2^{\frac{3}{4}}*2^{\frac{4}{3}}}{4^{\frac{3}{2}}}=\frac{2*2*2^{\frac{1}{12}}}{2*2*2}=\frac{2^{\frac{1}{12}}}{2}=2^{\frac{1}{12}-\frac{12}{12}}=2^{-\frac{11}{12}}=\frac{1}{2^{\frac{11}{12}}}=\frac{1}{\sqrt[12]{2^{11}}}=\frac{1}{(\sqrt[12]{2})^{11}}[/tex]
[tex]\frac{2^{\frac{3}{4}}*2^{\frac{4}{3}}}{4^{\frac{3}{2}}}[/tex]
Teller:
[tex]2^{\frac{3}{4}}*2^{\frac{4}{3}}=\sqrt[4]{2^{3}}*\sqrt[3]{2^{4}}=2^{\frac{3}{4}+\frac{4}{3}}=2^{\frac{9}{12}+\frac{16}{12}}=2^{\frac{25}{12}}=2^{\frac{12}{12}}*2^{\frac{12}{12}}*2^{\frac{1}{12}}=2*2*2^{\frac{1}{12}}[/tex]
Nevner:
[tex]4^{\frac{3}{2}}=\sqrt{4^{3}}=\sqrt{64}=8=2*2*2[/tex]
Brøk:
[tex]\frac{2^{\frac{3}{4}}*2^{\frac{4}{3}}}{4^{\frac{3}{2}}}=\frac{2*2*2^{\frac{1}{12}}}{2*2*2}=\frac{2^{\frac{1}{12}}}{2}=2^{\frac{1}{12}-\frac{12}{12}}=2^{-\frac{11}{12}}=\frac{1}{2^{\frac{11}{12}}}=\frac{1}{\sqrt[12]{2^{11}}}=\frac{1}{(\sqrt[12]{2})^{11}}[/tex]