matematikk.net

a^2/3*(b^3/2)-1
3 √ a* √ b

Kan noen hjelpe mej?

a[sup](2/3)[/sup] * b[sup](3/2)[/sup] -1
-------------------------------
3 [symbol:rot](a) * [symbol:rot](b)


a[sup](2/3)[/sup] * b[sup](3/2)[/sup] -1
--------------------------------
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]
------------------------------------------------
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]]
--------------------------------------------------------------------------
3

[a[sup](1/6)[/sup]*b]-[a[sup](-1/2)[/sup] * b[sup](-1/2)[/sup]]
------------------------------------------------------
3

Svaret skal bli:

a[sup]1/3[/sup]
b[sup]2[/sup]

Stemmer ditt svar da?:/

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]

kult