# Simplify the aritmetic expression: [3/4 ·1/4 ·(5− 3/2)-: (3/4 − 3/16)] -: 7/4 ·(2 + 1/2)^2 −(1 + 1/2)^2?

Apr 28, 2016

$\frac{23}{12}$

#### Explanation:

Given,

$\left[\frac{3}{4} \cdot \frac{1}{4} \cdot \left(5 - \frac{3}{2}\right) \div \left(\frac{3}{4} - \frac{3}{16}\right)\right] \div \frac{7}{4} \cdot {\left(2 + \frac{1}{2}\right)}^{2} - {\left(1 + \frac{1}{2}\right)}^{2}$

According to B.E.D.M.A.S., start by simplifying the round bracketed terms in the square brackets.

$= \left[\frac{3}{4} \cdot \frac{1}{4} \cdot \left(\textcolor{b l u e}{\frac{10}{2}} - \frac{3}{2}\right) \div \left(\textcolor{b l u e}{\frac{12}{16}} - \frac{3}{16}\right)\right] \div \frac{7}{4} \cdot {\left(2 + \frac{1}{2}\right)}^{2} - {\left(1 + \frac{1}{2}\right)}^{2}$

$= \left[\frac{3}{4} \cdot \frac{1}{4} \cdot \left(\textcolor{b l u e}{\frac{7}{2}}\right) \div \left(\textcolor{b l u e}{\frac{9}{16}}\right)\right] \div \frac{7}{4} \cdot {\left(2 + \frac{1}{2}\right)}^{2} - {\left(1 + \frac{1}{2}\right)}^{2}$

Omit the round brackets in the square brackets.

$= \left[\frac{3}{4} \cdot \frac{1}{4} \cdot \frac{7}{2} \div \frac{9}{16}\right] \div \frac{7}{4} \cdot {\left(2 + \frac{1}{2}\right)}^{2} - {\left(1 + \frac{1}{2}\right)}^{2}$

Simplify the expression within the square brackets.

$= \left[\frac{3}{16} \cdot \frac{7}{2} \div \frac{9}{16}\right] \div \frac{7}{4} \cdot {\left(2 + \frac{1}{2}\right)}^{2} - {\left(1 + \frac{1}{2}\right)}^{2}$

$= \left[\frac{21}{32} \cdot \frac{16}{9}\right] \div \frac{7}{4} \cdot {\left(2 + \frac{1}{2}\right)}^{2} - {\left(1 + \frac{1}{2}\right)}^{2}$

$= \left[\frac{21 \textcolor{red}{\div 3}}{32 \textcolor{p u r p \le}{\div 16}} \cdot \frac{16 \textcolor{p u r p \le}{\div 16}}{9 \textcolor{red}{\div 3}}\right] \div \frac{7}{4} \cdot {\left(2 + \frac{1}{2}\right)}^{2} - {\left(1 + \frac{1}{2}\right)}^{2}$

$= \left[\frac{7}{2} \cdot \frac{1}{3}\right] \div \frac{7}{4} \cdot {\left(2 + \frac{1}{2}\right)}^{2} - {\left(1 + \frac{1}{2}\right)}^{2}$

$= \left[\frac{7}{6}\right] \div \frac{7}{4} \cdot {\left(2 + \frac{1}{2}\right)}^{2} - {\left(1 + \frac{1}{2}\right)}^{2}$

Omit the square brackets since the term is already simplified.

$= \frac{7}{6} \div \frac{7}{4} \cdot {\left(2 + \frac{1}{2}\right)}^{2} - {\left(1 + \frac{1}{2}\right)}^{2}$

Continue simplifying the terms in the round brackets.

$= \frac{7}{6} \div \frac{7}{4} \cdot {\left(\frac{4}{2} + \frac{1}{2}\right)}^{2} - {\left(\frac{2}{2} + \frac{1}{2}\right)}^{2}$

$= \frac{7}{6} \div \frac{7}{4} \cdot {\left(\frac{5}{2}\right)}^{2} - {\left(\frac{3}{2}\right)}^{2}$

$= \frac{7}{6} \div \frac{7}{4} \cdot \left(\frac{25}{4}\right) - \left(\frac{9}{4}\right)$

Omit the round brackets since the bracketed terms are already simplified.

$= \frac{7}{6} \div \frac{7}{4} \cdot \frac{25}{4} - \frac{9}{4}$

$= \frac{7}{6} \cdot \frac{4}{7} \cdot \frac{25}{4} - \frac{9}{4}$

The $7$'s and $4$'s cancel each other out since they appear in the numerator and denominator as a pair.

$= \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}}{6} \cdot \frac{\textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{4}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}} \cdot \frac{25}{\textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{4}}}} - \frac{9}{4}$

$= \frac{25}{6} - \frac{9}{4}$

Change the denominator of each fraction such that both fractions have the same denominator.

$= \frac{25}{\textcolor{red}{6}} \left(\frac{\textcolor{p u r p \le}{4}}{\textcolor{p u r p \le}{4}}\right) - \frac{9}{\textcolor{p u r p \le}{4}} \left(\frac{\textcolor{red}{6}}{\textcolor{red}{6}}\right)$

$= \frac{100}{24} - \frac{54}{24}$

$= \frac{46}{24}$

$= \frac{23}{12}$