Let's simplify and solve the given mathematical problems step by step.
2.1 Simplification
2.1.1 Simplify:
\[ \frac{2x - 4}{x + 1} \cdot \frac{4}{x + 2} \cdot \frac{2x - 1}{4} \]
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Factor where possible: \[ 2x - 4 = 2(x - 2) \] So the expression becomes: \[ \frac{2(x - 2)}{x + 1} \cdot \frac{4}{x + 2} \cdot \frac{2x - 1}{4} \]
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Cancel \(4\) in the numerator and denominator: \[ = \frac{2(x - 2)(2x - 1)}{(x + 1)(x + 2)} \]
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Final result: \[ \frac{2(x - 2)(2x - 1)}{(x + 1)(x + 2)} \]
2.1.2 Simplify:
\[ 2^8 - 4 \cdot 2^5 + 3 \cdot 2^0 \]
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Calculate \(4\) and \(3\) as powers of \(2\): \[ 4 = 2^2, \quad 2^5 = 32, \quad 3 \cdot 2^0 = 3 \]
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Plug back into the expression: \[ = 2^8 - 2^2 \cdot 2^5 + 3 \] \[ = 2^8 - 2^{2+5} + 3 = 2^8 - 2^7 + 3 \]
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Combine the first two terms: \[ = 2^7(2 - 1) + 3 = 2^7 + 3 = 128 + 3 = 131 \]
2.2 Solve for \(x\)
2.2.1 Solve:
\[ x^{-\frac{3}{4}} = 27 \]
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Rewrite \(27\) as \(3^3\): \[ x^{-\frac{3}{4}} = 3^3 \]
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Take the reciprocal and adjust the exponent: \[ x^{\frac{3}{4}} = \frac{1}{3^3} = \frac{1}{27} \]
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Raise both sides to the \(\frac{4}{3}\): \[ x = \left(\frac{1}{27}\right)^{\frac{4}{3}} = \frac{1^4}{27^{\frac{4}{3}}} = \frac{1}{729} \text{ (since } 27^{\frac{4}{3}} = 3^{4} = 81\text{)} \]
2.2.2 Solve:
\[ 2^{x} - \frac{1}{2} = 0 \]
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Add \(\frac{1}{2}\) to both sides: \[ 2^{x} = \frac{1}{2} \]
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Rewrite \(\frac{1}{2}\) as \(2^{-1}\): \[ 2^{x} = 2^{-1} \quad \Rightarrow \quad x = -1 \]
2.2.3 Solve:
\[ 2^{2x + 3} \cdot 2^{x - 4} = 0 \]
- Combine exponents: \[ 2^{(2x + 3) + (x - 4)} = 0 \] This implies \(2^{3x - 1} = 0\), which cannot occur for any real \(x\).
3.1 Linear Pattern Analysis
3.1.1 Find the formula for the nth term:
Given the sequence: 9, 5, 1, ..., -143.
- The first term (\(a_1\)) is \(9\) and common difference (\(d = 5 - 9 = -4\)).
- The \(n\)th term formula: \[ T_n = a_1 + (n - 1)d = 9 + (n - 1)(-4) = 9 - 4(n - 1) = -4n + 13 \]
3.1.2 Calculate \(T_7\):
\[ T_7 = -4(7) + 13 = -28 + 13 = -15 \]
3.1.3 Determine the number of terms in the pattern:
Set \(T_n = -143\): \[ -4n + 13 = -143 \] Solving for \(n\): \[ -4n = -143 - 13 \quad \Rightarrow \quad -4n = -156 \quad \Rightarrow \quad n = 39 \]
3.2 Linear Number Pattern Analysis
The first three terms are:
- \(2x - 1\)
- \(4x - 5\)
- \(3x + 3\)
Set the terms equal:
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\(4x - 5 - (2x - 1) = -3\) \[ 2x - 4 = -3 \quad \Rightarrow \quad 2x = 1 \quad \Rightarrow \quad x = \frac{1}{2} \]
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First three terms: \[ 2(\frac{1}{2}) - 1 = 0, \quad 4(\frac{1}{2}) - 5 = -3, \quad 3(\frac{1}{2}) + 3 = 4.5 \] Which leads to:
- First term = 0
- Second term = -3
- Third term = 4.5
Final Answers
2.1.1: \(\frac{2(x - 2)(2x - 1)}{(x + 1)(x + 2)}\)
2.1.2: 131
2.2.1: \(x = \frac{1}{729}\)
2.2.2: \(x = -1\)
2.2.3: No solution (as it results in an impossible equation).
3.1.1: \(T_n = -4n + 13\)
3.1.2: \(T_7 = -15\)
3.1.3: \(39\)
3.2.1: \(x = \frac{1}{2}\)
3.2.2: First terms are \(0, -3, 4.5\) respectively.
1 answer