Step 1: Read the function.
We have $f(x)=\dfrac{1}{\sqrt{|x|-[x]}}$, where $[x]$ is the floor of $x$. The square root sits in the denominator, so whatever is inside must be strictly greater than zero.
Step 2: Set up the domain condition.
We need $|x|-[x]>0$. Let us think about this for $x\ge 0$ first, since the positive side is what controls the answer.
Step 3: Simplify on the non-negative side.
For $x\ge 0$ we have $|x|=x$, so the quantity becomes $x-[x]=\{x\}$, the fractional part of $x$. This is positive exactly when $x$ is not an integer, and it is zero when $x$ is a whole number.
Step 4: Describe the domain A.
So on the positive side every $x>0$ that is not a positive integer works, and $x=0$ fails because the inside becomes $0$. Hence the domain restricted here is $A=R^{+}-Z^{+}$, that is, all positive reals except the positive integers.
Step 5: Describe the range B.
Since $0<\{x\}<1$, the inside $|x|-[x]$ takes values in $(0,1)$, so $\sqrt{|x|-[x]}\in(0,1)$, and its reciprocal $f(x)=\dfrac{1}{\sqrt{\{x\}}}$ is greater than $1$. Thus the output is always a positive real number, so the range $B=R^{+}$.
Step 6: Take the intersection.
We need $A\cap B=(R^{+}-Z^{+})\cap R^{+}$. Every element of $A$ is already a positive real, so the intersection is just $A$ itself, namely $R^{+}-Z^{+}$. That matches option (1).
\[ \boxed{A\cap B=R^{+}-Z^{+}} \]