1/ $\displaystyle 3\sqrt{(x^3y^4)^2}$

2/ $\displaystyle \frac{(x^2y^4)^2}{x^5y^-1}$

3/ (x^3a+1)(x^a-3)/4sqrtx^8

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- Jan 14th 2009, 12:27 PMwilliamsimplify
1/ $\displaystyle 3\sqrt{(x^3y^4)^2}$

2/ $\displaystyle \frac{(x^2y^4)^2}{x^5y^-1}$

3/ (x^3a+1)(x^a-3)/4sqrtx^8 - Jan 14th 2009, 12:38 PMmasters
Hello william,

[1] $\displaystyle 3\sqrt{(x^3y^4)^2}=3x^3y^4$

[2] $\displaystyle \frac{(x^2y^4)^2}{x^5y^-1}=\frac{x^4y^8y}{x^5}=\frac{x^9}{x}$

[3] $\displaystyle \frac{(x^{3a}+1)(x^a-3)}{4\sqrt{x^8}}=\frac{x^{4a}-3x^{3a}+x^{a}-3}{4x^4}$

I made a few corrections on [3]. I'm not sure I translated your 3rd one correctly. Let me know. - Jan 14th 2009, 12:43 PMwilliam
Thanks alot

What did you do there for 1, I am assuming you just wrote the second question as the first by mistake. For 3,

For 2 how did you get y^8 and not y^16, can you show steps please?

$\displaystyle \frac{(x^{3a+1})(x^{a-3}}{4\sqrt{x^8}}$

in the numerator it is x^{3a+1} and x^{a-3}

and if you don't mind me asking how do i simplify (27)^-2/3 - Jan 14th 2009, 01:01 PMmasters
- Jan 14th 2009, 01:02 PMwilliam
- Jan 14th 2009, 01:14 PMmasters
William,

Do you know that $\displaystyle \sqrt{a^2}=a$?

That's what I used to answer the first one. You could also look at it this way:

$\displaystyle \sqrt{a}=\sqrt{a^1}=a^{1/2}$. You can convert this into rational exponents by dividing the power in the radicand by the index of the root.

$\displaystyle 3\sqrt{(x^3y^4)^2}=3(x^3y^4)^{2/2}=3(x^3 y^4)^1=3x^3 y^4$

Or, you could expand the radicand first and then simplify at the end like

$\displaystyle 3 \sqrt{(x^3y^4)^2}=3\sqrt{x^6y^8}=3x^{6/2}y^{8/2}=3x^3y^4$ - Jan 14th 2009, 01:19 PMwilliam
- Jan 14th 2009, 01:37 PMmasters
Use this property of exponents: $\displaystyle (a^m)^n=a^{mn}$. So $\displaystyle (y^4)^2=y^{4 \cdot 2}=y^8$

Is it $\displaystyle \frac{(x^{3a+1})(x^{a-3})}{4\sqrt{x^8}}$

If so, $\displaystyle \frac{(x^{3a+1})(x^{a-3})}{4\sqrt{x^8}}=\frac{x^{(3a+3)+(a-3)}}{4x^4}=\frac{x^{4a}}{4x^4}=\frac{x^{4a-4}}{4}=\frac{x^{4(a-1)}}{4}$

$\displaystyle 27^{-\frac{2}{3}}=\frac{1}{27^{\frac{2}{3}}}=\frac{1}{( \sqrt[3]{27})^2}=\frac{1}{(3)^2}=\frac{1}{9}$ - Jan 14th 2009, 01:42 PMwilliam