1. **State the problem:** Simplify the expression $$\frac{5}{4} - 3\left[ \left( 1 - \frac{1}{4} \right)^{-1} + \frac{2}{3} \left( \frac{1}{2} - \frac{5}{4} \right) - \frac{3}{4} : \frac{6}{2} \right]$$ 2. **Recall important rules:** - The inverse power $x^{-1}$ means $\frac{1}{x}$. - Division $a : b$ is the same as $\frac{a}{b}$. - Follow order of operations: parentheses, exponents, multiplication/division, addition/subtraction. 3. **Simplify inside the brackets step-by-step:** - Calculate $1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}$. - Then $\left( \frac{3}{4} \right)^{-1} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$. - Calculate $\frac{1}{2} - \frac{5}{4} = \frac{2}{4} - \frac{5}{4} = -\frac{3}{4}$. - Multiply by $\frac{2}{3}$: $\frac{2}{3} \times -\frac{3}{4} = -\frac{6}{12} = -\frac{1}{2}$. - Division $\frac{3}{4} : \frac{6}{2} = \frac{3}{4} \times \frac{2}{6} = \frac{3 \times 2}{4 \times 6} = \frac{6}{24} = \frac{1}{4}$. 4. **Sum inside the brackets:** $$\frac{4}{3} + \left(-\frac{1}{2}\right) - \frac{1}{4} = \frac{4}{3} - \frac{1}{2} - \frac{1}{4}$$ Find common denominator 12: $$\frac{4}{3} = \frac{16}{12}, \quad \frac{1}{2} = \frac{6}{12}, \quad \frac{1}{4} = \frac{3}{12}$$ So sum is: $$\frac{16}{12} - \frac{6}{12} - \frac{3}{12} = \frac{16 - 6 - 3}{12} = \frac{7}{12}$$ 5. **Multiply by 3:** $$3 \times \frac{7}{12} = \frac{21}{12} = \frac{7}{4}$$ 6. **Final subtraction:** $$\frac{5}{4} - \frac{7}{4} = -\frac{2}{4} = -\frac{1}{2}$$ **Final answer:** $$-\frac{1}{2}$$