i dont understand what to do when is it like this
(9+16)^1/2
my answer was 7? is that correct???
and what if it is like this
9^1/2 + 16^1/2
i also got 7???
i have more questions...
When dividing you subtract the indices and when multiplying you add the indices (first the indices need to have a common denominator):
$\displaystyle x^{\frac{5}{8}}\div x^{\frac{1}{6}}= x^{\frac{15}{24}}\div x^{\frac{4}{24}}$$\displaystyle =x^{\frac{11}{24}}$
For the second one you need to add the indices:
$\displaystyle x^{\frac{2}{3}}(x^{\frac{1}{4}}-x)= x^{\frac{8}{12}}\times x^{\frac{3}{12}}-x^{\frac{2}{3}}\times x^{\frac{3}{3}}=x^{\frac{11}{12}}-x^{\frac{5}{3}}$
Does that make sense?
For the second one you need to add the indices:
$\displaystyle x^{\frac{2}{3}}(x^{\frac{1}{4}}-x)= x^{\frac{8}{12}}\times x^{\frac{3}{12}}-x^{\frac{2}{3}}\times x^{\frac{3}{3}}=x^{\frac{11}{12}}-x^{\frac{5}{3}}$
Does that make sense?[/quote]
im a little confused how do you get x^11/12?
$\displaystyle x^{\frac{2}{3}}(x^{\frac{1}{4}}-x)$ is the same as $\displaystyle x^{\frac{8}{12}}(x^{\frac{3}{12}}-x)$
So as you're multiplying $\displaystyle x^{\frac{8}{12}}$ by $\displaystyle x^{\frac{3}{12}}$ the indices added together $\displaystyle =\frac{11}{12}$
Similarly, as $\displaystyle x^{\frac{2}{3}}(x^{\frac{1}{4}}-x)$ is the same as $\displaystyle x^{\frac{2}{3}}(x^{\frac{1}{4}}-x^{\frac{3}{3}})$ you get $\displaystyle x^{\frac{5}{3}}$ as the second term.
$\displaystyle (x^{\frac{1}{3}}+x^{-\frac{1}{3}})^2$
$\displaystyle =\left( x^{\frac{1}{3}}+\frac{1}{x^{\frac{1}{3}}}\right)\l eft( x^{\frac{1}{3}}+\frac{1}{x^{\frac{1}{3}}}\right)$
$\displaystyle =(x^{\frac{1}{3}})^2+2(x^{\frac{1}{3}})(x^{-\frac{1}{3}})+(x^{-\frac{1}{3}})^2$
$\displaystyle =x^{\frac{2}{3}}+2+\frac{1}{x^{\frac{2}{3}}}$
Nope. That's fully simplified.
Your answer is correct. $\displaystyle \sqrt[3]{125}=5$ as $\displaystyle 5\times 5\times 5=125$
You should check out the exponent section in this site Practical Algebra Lessons , it explains them really well.
Whoa, that's not right. You have to FOIL it:
$\displaystyle (x^{\frac{1}{3}}+x^{-\frac{1}{3}})^2$
$\displaystyle \begin{aligned}
&= (x^{\frac{1}{3}})^2 + 2(x^{\frac{1}{3}})(x^{-\frac{1}{3}}) + (x^{-\frac{1}{3}})^2 \\
&= x^{\frac{2}{3}} + 2 + x^{-\frac{2}{3}}
\end{aligned}$
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