Question

Okay, help me simplify an expression! Come to the question so it can be displayed clearly as equation. Okay, help me simplify an expression! Come to the question so it can be displayed clearly as equation. @Mathematics

Answer 1

\[(x^6y^4)^(1/8)+2(x^(1/3)y^(1/4))^2\]

Answer 2

\[(x^6y^4)^{1/8}+2(x^{1/3}y^{1/4})^2\]is written into the equation feature as

(x^6y^4)^{1/8}+2(x^{1/3}y^{1/4})^2

note the brackets on the exponents. just so you can write it correctly in the future. Ok, your problem...

Answer 3

distribute the exponents:\[(x^6y^4)^{1/8}+2(x^{1/3}y^{1/4})^2=x^{3/4}y^{1/2}+2x^{2/3}y^{1/2}\]factor out the y^(1/2)\[=y^{1/2}(x^{3/4}+2x^{2/3})\]

Answer 4

remember
\[(x^a)^b=x^{ab}\]

Answer 5

Thank you! ^_^ You're very helpful!

Answer 6

Oh but do you think you can help me with another? I'm sorry, but if you can herre it is:
\[(3\sqrt[3]{7^3}+4\sqrt[3]{7^3})/\sqrt{7^5}\]

Answer 7

\[{3\sqrt[3]{7^3}+4\sqrt[3]{7^3}\over \sqrt{7^5}}={3(7^{3/3})+4(7^{3/3})\over7^{5/2}}={3(7)+4(7)\over7^27^{1/2}}\]\[={7(3+4)\over7^27^{1/2}}={7^2\over7^27^{1/2}}={1\over \sqrt7}\]

Answer 8

you can write the final answer also as\[{1\over \sqrt7}={\sqrt7\over7}={1\over7^{1/2}}=7^{-1/2}\]all these are equivalent, so take your pick.

Answer 9

:O :D Well thanks!!! You're really awesome! Thank you thank youu! :) Thats all my math for today, :)))

Answer 10

:) anytime!