Hello Punch Originally Posted by
Punch Simplify $\displaystyle (x+2)[\frac{\frac{-3}{2}}{(3x+1)^(1.5)}]+\frac{1}{\sqrt{3x+1}}$
>>>>;;;;Note that 1.5 is a power of 3x+1;;;;<<<<
If I re-write the question, I think it is: Simplify:$\displaystyle (x+2)\left(\frac{-\frac{3}{2}}{(3x+1)^{1.5}}\right)+\frac{1}{\sqrt{3 x+1}}$
First, note that the $\displaystyle (x+2)$ is multiplying the first fraction, so it goes on top; the $\displaystyle -\tfrac32$ can be written with the $\displaystyle -3$ on top and the $\displaystyle 2$ underneath. So we get:$\displaystyle =\frac{-3(x+2)}{2(3x+1)^{1.5}}+\frac{1}{\sqrt{3x+1}}$
Do you see how that works?
Then $\displaystyle (...)^{1.5} = (...)^1\times(...)^{0.5}=(...)\sqrt{(...)}$.
So the next stage is: $\displaystyle =\frac{-3(x+2)}{2(3x+1)\sqrt{3x+1}}+\frac{1}{\sqrt{3x+1}}$
OK. Now find the lowest common denominator, which is $\displaystyle 2(3x+1)\sqrt{3x+1}$. So, writing the second fraction with this denominator, we get:$\displaystyle =\frac{-3(x+2)}{2(3x+1)\sqrt{3x+1}}+\frac{2(3x+1)}{2(3x+1) \sqrt{3x+1}}$
Now combine the fractions over a single denominator and simplify:$\displaystyle =\frac{-3(x+2)+2(3x+1)}{2(3x+1)\sqrt{3x+1}}$
$\displaystyle =\frac{3x-4}{2(3x+1)\sqrt{3x+1}}$
which you can write as:$\displaystyle =\frac{3x-4}{2(3x+1)^{1.5}}$
if you like.
Grandad