Simplify $\left(\dfrac{2^{a}-\sqrt{2}}{2^{a}}\right)^{2}$. If $27m^{3x}=1$ and $m>0$, find $m^{2x}+m^{-x}$.

1. Separate the fraction and simplify each term using indices

$\left(\dfrac{2^{a}-\sqrt{2}}{2^{a}}\right)^2=\left(\dfrac{2^a}{2^a}-\dfrac{\sqrt2}{2^a}\right)^2$
$=\left(1-\dfrac{2^{\frac12}}{2^a}\right)^2$
$=\left(1-2^{\frac12-a}\right)^2$

2. Expand the square and simplify powers of 2

$\left(1-2^{\frac12-a}\right)^2=1-2\left(2^{\frac12-a}\right)+\left(2^{\frac12-a}\right)^2$
$=1-2^{1+\frac12-a}+2^{2\left(\frac12-a\right)}$
$=1-2^{\frac32-a}+2^{1-2a}$

1. Express the given equation using $m^x$ and solve for $m^x$

$27m^{3x}=1$
$m^{3x}=\frac{1}{27}=\frac{1}{3^3}$
Since $m>0$, $m^x>0$.
$(m^x)^3=\left(\frac{1}{3}\right)^3$
$m^x=\frac{1}{3}$

2. Substitute $m^x=\frac13$ into the required expression

$m^{2x}+m^{-x}=(m^x)^2+\frac{1}{m^x}$
$=\left(\frac{1}{3}\right)^2+\frac{1}{\frac{1}{3}}$
$=\frac{1}{9}+3$
$=\frac{28}{9}$