Let's simplify and solve the given mathematical problems step by step.
### 2.1 Simplification
#### 2.1.1 Simplify:
\[
\frac{2x - 4}{x + 1} \cdot \frac{4}{x + 2} \cdot \frac{2x - 1}{4}
\]
1. Factor where possible:
\[
2x - 4 = 2(x - 2)
\]
So the expression becomes:
\[
\frac{2(x - 2)}{x + 1} \cdot \frac{4}{x + 2} \cdot \frac{2x - 1}{4}
\]
2. Cancel \(4\) in the numerator and denominator:
\[
= \frac{2(x - 2)(2x - 1)}{(x + 1)(x + 2)}
\]
3. Final result:
\[
\frac{2(x - 2)(2x - 1)}{(x + 1)(x + 2)}
\]
#### 2.1.2 Simplify:
\[
2^8 - 4 \cdot 2^5 + 3 \cdot 2^0
\]
1. Calculate \(4\) and \(3\) as powers of \(2\):
\[
4 = 2^2, \quad 2^5 = 32, \quad 3 \cdot 2^0 = 3
\]
2. Plug back into the expression:
\[
= 2^8 - 2^2 \cdot 2^5 + 3
\]
\[
= 2^8 - 2^{2+5} + 3 = 2^8 - 2^7 + 3
\]
3. 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:
\[
x^{-\frac{3}{4}} = 27
\]
1. Rewrite \(27\) as \(3^3\):
\[
x^{-\frac{3}{4}} = 3^3
\]
2. Take the reciprocal and adjust the exponent:
\[
x^{\frac{3}{4}} = \frac{1}{3^3} = \frac{1}{27}
\]
3. 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:
\[
2^{x} - \frac{1}{2} = 0
\]
1. Add \(\frac{1}{2}\) to both sides:
\[
2^{x} = \frac{1}{2}
\]
2. Rewrite \(\frac{1}{2}\) as \(2^{-1}\):
\[
2^{x} = 2^{-1} \quad \Rightarrow \quad x = -1
\]
#### 2.2.3 Solve:
\[
2^{2x + 3} \cdot 2^{x - 4} = 0
\]
1. 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.
1. The first term (\(a_1\)) is \(9\) and common difference (\(d = 5 - 9 = -4\)).
2. 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:
1. \(2x - 1\)
2. \(4x - 5\)
3. \(3x + 3\)
Set the terms equal:
1. \(4x - 5 - (2x - 1) = -3\)
\[
2x - 4 = -3 \quad \Rightarrow \quad 2x = 1 \quad \Rightarrow \quad x = \frac{1}{2}
\]
2. 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.