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
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.
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
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
$\displaystyle 3 \sqrt{(x^3y^4)^2}=3\sqrt{x^6y^8}=3x^{6/2}y^{8/2}=3x^3y^4$
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, $\displaystyle \frac{(x^{3a+1})(x^{a-3})}{4\sqrt{x^8}}=\frac{x^{(3a+3)+(a-3)}}{4x^4}=\frac{x^{4a}}{4x^4}=\frac{x^{4a-4}}{4}=\frac{x^{4(a-1)}}{4}$
$\displaystyle 27^{-\frac{2}{3}}=\frac{1}{27^{\frac{2}{3}}}=\frac{1}{( \sqrt[3]{27})^2}=\frac{1}{(3)^2}=\frac{1}{9}$