a^2/3*(b^3/2)-1
3 √ a* √ b
Kan noen hjelpe mej?
Algebra, treng hjelp!
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
a[sup](2/3)[/sup] * b[sup](3/2)[/sup] -1
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3 [symbol:rot](a) * [symbol:rot](b)
a[sup](2/3)[/sup] * b[sup](3/2)[/sup] -1
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3 * a[sup](1/2)[/sup] * b[sup](1/2)[/sup]
{a[sup](2/3)[/sup] * b[sup](3/2)[/sup] -1}*a[sup](-1/2)[/sup] * b[sup](-1/2)[/sup]
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3
[a[sup](2/3)[/sup] * b[sup](3/2)[/sup]*a[sup](-1/2)[/sup] * b[sup](-1/2)[/sup]]-[a[sup](-1/2)[/sup] * b[sup](-1/2)[/sup]]
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3
[a[sup](1/6)[/sup]*b]-[a[sup](-1/2)[/sup] * b[sup](-1/2)[/sup]]
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3
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3 [symbol:rot](a) * [symbol:rot](b)
a[sup](2/3)[/sup] * b[sup](3/2)[/sup] -1
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3 * a[sup](1/2)[/sup] * b[sup](1/2)[/sup]
{a[sup](2/3)[/sup] * b[sup](3/2)[/sup] -1}*a[sup](-1/2)[/sup] * b[sup](-1/2)[/sup]
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3
[a[sup](2/3)[/sup] * b[sup](3/2)[/sup]*a[sup](-1/2)[/sup] * b[sup](-1/2)[/sup]]-[a[sup](-1/2)[/sup] * b[sup](-1/2)[/sup]]
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3
[a[sup](1/6)[/sup]*b]-[a[sup](-1/2)[/sup] * b[sup](-1/2)[/sup]]
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3
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Solar Plexsus
- Over-Guru
- Innlegg: 1685
- Registrert: 03/10-2005 12:09
Ut fra fasitsvaret kan jeg tenke meg at det riktige algebraiske uttrykket er
[tex]\frac{a^{2/3} \: (b^{3/2})^{-1}}{\sqrt[3]{a} \: sqrt{b}} \;=\; \frac{a^{2/3} \: b^{-3/2}}{a^{1/3} \: b^{1/2}} \;=\; a^{2/3 - 1/3} \, b^{-3/2 - 1/2} \;=\; a^{1/3} \: b^{-2} \;=\; \frac{a^{1/3}}{b^2}.[/tex]
[tex]\frac{a^{2/3} \: (b^{3/2})^{-1}}{\sqrt[3]{a} \: sqrt{b}} \;=\; \frac{a^{2/3} \: b^{-3/2}}{a^{1/3} \: b^{1/2}} \;=\; a^{2/3 - 1/3} \, b^{-3/2 - 1/2} \;=\; a^{1/3} \: b^{-2} \;=\; \frac{a^{1/3}}{b^2}.[/tex]