Question

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

Answer 1

Given expression:
$$\begin{gathered} =9\frac{3}{4}÷\left[2\frac{1}{6}+\left\{4\frac{1}{3}-\left(1\frac{1}{2}+1\frac{3}{4}\right)\right\}\right] \\ =\frac{39}{4}÷\left[\frac{13}{6}+\left\{\frac{13}{3}-\left(\frac{3}{2}+\frac{7}{4}\right)\right\}\right] \\ =\frac{39}{4}÷\left[\frac{13}{6}+\left\{\frac{13}{3}-\left(\frac{6+7}{4}\right)\right\}\right] \\ =\frac{39}{4}÷\left[\frac{13}{6}+\left\{\frac{13}{3}-\frac{13}{4}\right\}\right]\left(\mathrm{Removing}\mathrm{parentheses}\right) \\ =\frac{39}{4}÷\left[\frac{13}{6}+\left\{\frac{52-39}{12}\right\}\right] \\ =\frac{39}{4}÷\left[\frac{13}{6}+\frac{13}{12}\right]\left(\mathrm{Removing}\mathrm{braces}\right) \\ =\frac{39}{4}÷\left[\frac{26+13}{12}\right] \\ =\frac{39}{4}÷\frac{39}{12}(\mathrm{Removing}\mathrm{square}\mathrm{brackets}) \\ =\frac{39}{4}\times \frac{12}{39}=3(\mathrm{Removing}\text{'}÷\text{'}) \end{gathered}$$