$$9\frac{3}{4} \div \left[2\frac{1}{6} \div \left\{4\frac{1}{3} - \left(2\frac{1}{2} +\frac{3}{4} \right)\right\}\right]$$ is equal to:

From the given question,

$$9\frac{3}{4} \div \left[2\frac{1}{6} \div \left\{4\frac{1}{3} - \left(2\frac{1}{2} +\frac{3}{4} \right)\right\}\right]$$

Starts solving from small brackets (),

$$\Rightarrow 9\frac{3}{4} \div \left[2\frac{1}{6} \div \left\{4\frac{1}{3} - \left(\frac{5}{2} +\frac{3}{4} \right)\right\}\right]$$

$$\Rightarrow 9\frac{3}{4} \div \left[2\frac{1}{6} \div \left\{4\frac{1}{3} - \left(\frac{10+3}{4}  \right)\right\}\right]$$

$$\Rightarrow 9\frac{3}{4} \div \left[2\frac{1}{6} \div \left\{4\frac{1}{3} - \frac{13}{4}\right\}\right]$$

Start solving middle brackets {}

$$\Rightarrow 9\frac{3}{4} \div \left[2\frac{1}{6} \div \left\{\frac{13}{3} - \frac{13}{4}\right\}\right]$$

$$\Rightarrow 9\frac{3}{4} \div \left[2\frac{1}{6} \div \left\{\frac{13\times 4-13\times3}{3\times4} \right\}\right]$$

$$\Rightarrow 9\frac{3}{4} \div \left[2\frac{1}{6} \div \left\{\frac{52-39}{3\times4} \right\}\right]$$

$$\Rightarrow 9\frac{3}{4} \div \left[2\frac{1}{6} \div \frac{13}{12} \right]$$

Start solving large brackets []

$$\Rightarrow 9\frac{3}{4} \div \left[\frac{13}{6} \div \frac{13}{12} \right]$$

$$\Rightarrow 9\frac{3}{4} \div2$$

$$\Rightarrow \frac{39}{4} \div2$$

$$\Rightarrow \frac{39}{8}$$