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2.1 Simplify WITHOUT using a calculator: 2.1.1 \( \left(2^{-1}+3^{-1}\right)^{2} \) 2.1.2 \( \frac{2.4^{x+2}-4^{x-3.7}}{2^{x} \cdot 2^{x}} \) 2.1.3 \( 3^{-\frac{1}{2}}[\sqrt{12}+\sqrt[3]{3 \sqrt{3}}] \)

Ask by Pollard Weston. in South Africa
Feb 03,2025

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Tutor-Verified Answer

Answer

1. \( \frac{25}{36} \) 2. \( 32 - 2^{-7.4} \) 3. \( 3 \)

Solution

Let's simplify each expression step by step without using a calculator. --- ### **2.1.1 \( \left(2^{-1} + 3^{-1}\right)^{2} \)** 1. **Evaluate the Negative Exponents:** \[ 2^{-1} = \frac{1}{2}, \quad 3^{-1} = \frac{1}{3} \] 2. **Add the Fractions:** \[ \left(\frac{1}{2} + \frac{1}{3}\right) = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \] 3. **Square the Sum:** \[ \left(\frac{5}{6}\right)^2 = \frac{25}{36} \] **Simplified Form:** \[ \frac{25}{36} \] --- ### **2.1.2 \( \frac{2 \cdot 4^{x+2} - 4^{x-3.7}}{2^{x} \cdot 2^{x}} \)** 1. **Express All Terms with Base 2:** \[ 4 = 2^2 \Rightarrow 4^{x+2} = (2^2)^{x+2} = 2^{2x+4}, \quad 4^{x-3.7} = (2^2)^{x-3.7} = 2^{2x-7.4} \] 2. **Rewrite the Numerator:** \[ 2 \cdot 4^{x+2} - 4^{x-3.7} = 2 \cdot 2^{2x+4} - 2^{2x-7.4} = 2^{2x+5} - 2^{2x-7.4} \] 3. **Simplify the Denominator:** \[ 2^{x} \cdot 2^{x} = 2^{2x} \] 4. **Combine Numerator and Denominator:** \[ \frac{2^{2x+5} - 2^{2x-7.4}}{2^{2x}} = 2^5 - 2^{-7.4} = 32 - \frac{1}{2^{7.4}} \] **Simplified Form:** \[ 32 - 2^{-7.4} \] *(Alternatively, \(32 - \frac{1}{2^{7.4}}\))* --- ### **2.1.3 \( 3^{-\frac{1}{2}} \left[ \sqrt{12} + \sqrt[3]{3 \sqrt{3}} \right] \)** 1. **Simplify Each Component:** - \(3^{-\frac{1}{2}} = \frac{1}{\sqrt{3}}\) - \(\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}\) - \(\sqrt[3]{3 \sqrt{3}} = \sqrt[3]{3^{1} \cdot 3^{\frac{1}{2}}} = \sqrt[3]{3^{\frac{3}{2}}} = 3^{\frac{1}{2}} = \sqrt{3}\) 2. **Combine the Terms Inside the Brackets:** \[ 2\sqrt{3} + \sqrt{3} = 3\sqrt{3} \] 3. **Multiply by \( \frac{1}{\sqrt{3}} \):** \[ \frac{1}{\sqrt{3}} \cdot 3\sqrt{3} = 3 \] **Simplified Form:** \[ 3 \] --- **Final Answers:** 1. **2.1.1:** \(\dfrac{25}{36}\) 2. **2.1.2:** \(32 - 2^{-7.4}\) 3. **2.1.3:** \(3\)

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The Deep Dive

To simplify the expressions, let's break it down step by step: **2.1.1** Start with the expression inside the parentheses: \( 2^{-1} + 3^{-1} = \frac{1}{2} + \frac{1}{3} \). Finding a common denominator (which is 6), we have \( \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \). Now, square this result: \[ \left( \frac{5}{6} \right)^{2} = \frac{25}{36}. \] **2.1.2** First, rewrite \( 4^{x-3.7} \) as \( (2^{2})^{x-3.7} = 2^{2x - 7.4} \). The original expression becomes: \[ \frac{2.4^{x+2} - 2^{2x - 7.4}}{2^{x} \cdot 2^{x}} = \frac{2^{x+2} - 2^{2x - 7.4}}{2^{2x}}. \] This simplifies to \( \frac{2^{x+2}}{2^{2x}} - \frac{2^{2x - 7.4}}{2^{2x}} = 2^{2-x} - 2^{-7.4} = 2^{2-x} - \frac{1}{2^{7.4}} \). **2.1.3** Start with the term \( 3^{-\frac{1}{2}} = \frac{1}{\sqrt{3}} \). Now calculate \( \sqrt{12} = 2\sqrt{3} \) and \( \sqrt[3]{3\sqrt{3}} = 3^{\frac{2}{3}} \). Expressing these together gives: \[ \sqrt{12} + \sqrt[3]{3 \sqrt{3}} = 2\sqrt{3} + 3^{\frac{2}{3}}. \] Finally, multiply by \( \frac{1}{\sqrt{3}} \): \[ 3^{-\frac{1}{2}}(2\sqrt{3} + 3^{\frac{2}{3}}) = \frac{2\sqrt{3}}{\sqrt{3}} + \frac{3^{\frac{2}{3}}}{\sqrt{3}} = 2 + 3^{\frac{1}{3}}. \]

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