How des one simplify this?
(2ab)^-2/2(ab)^3
I am unsure if this correct,
I removed the brackets
-4a^-2b^-2/2a^-3b^-3
to make exponents positive i moved to the top
-4a^-2--3b^-2--3 +2
which is -4ab+2
Is this correct?
It's all right. For the equation I posted, you have 2(ab)^3 for the denominator. Using the law of operations the power of 3 only applies to a and b, not on the 2.
As for the fixed equation
$\displaystyle \frac{(2ab)^{-2}}{2(ab)^{-3}} $
= $\displaystyle (2ab)^{-2} \cdot \left(2 (ab)^{-3}\right)^{-1} $
= $\displaystyle 2^{-2}a^{-2}b^{-2} \cdot 2^{-1} (ab)^3 $
= $\displaystyle 2^{-2}a^{-2}b^{-2} \cdot 2^{-1} a^3 b^3 $
= $\displaystyle 2^{-3} ab $
= $\displaystyle \frac{ab}{8} $
Hello, kiwisa!
Simplify: .$\displaystyle \dfrac{(2ab)^{-2}}{2(ab)^{-3}} $
We have: .$\displaystyle \dfrac{(2ab)^{-2}}{2(ab)^{-3}} \;=\;\dfrac{2^{-2}\cdot a^{-2}\cdot b^{-2}} {2\cdot a^{-3}\cdot b^{-3}}$
Use the exponent rules: when dividing, subtract exponents.
And we have: .$\displaystyle 2^{-2-1}\cdot a^{-2-(-3)}\cdot b^{-1-(-3)}$
. . . . . . . . $\displaystyle =\;2^{-3}\cdot a^1\cdot b^1$
. . . . . . . . $\displaystyle =\; \dfrac{ab}{2^3}$
. . . . . . . . $\displaystyle =\;\dfrac{ab}{8}$