https://r2.erweima.ai/mathbot/image/17265022079047xhrlfy8.jpg

Let's analyze and solve the four problems presented in the image:

### 1. \( 5x + 7 \equiv (m + n)x + 2m - 1 \)
We are asked to calculate the value of \( m - n \).

- Expand both sides:
  \[
  5x + 7 = (m+n)x + 2m - 1
  \]
  
  Compare the coefficients of \( x \) on both sides:
  \[
  5 = m + n
  \]
  
  And compare the constant terms:
  \[
  7 = 2m - 1
  \]
  
  Solve for \( m \) and \( n \):
  - From the constant term equation \( 7 = 2m - 1 \):
    \[
    7 + 1 = 2m \implies m = 4
    \]
  - Substitute \( m = 4 \) into \( 5 = m + n \):
    \[
    5 = 4 + n \implies n = 1
    \]
  
  Therefore, \( m - n = 4 - 1 = 3 \).

### 2. \( A(x - 2) + B(x + 3) \equiv 6x + B \)
We are asked to determine the value of \( A \times B \).

- Expand both sides:
  \[
  A(x - 2) + B(x + 3) = A(x) - 2A + B(x) + 3B = (A + B)x + (-2A + 3B)
  \]
  
  Now compare the coefficients of \( x \):
  \[
  A + B = 6
  \]
  
  And compare the constant terms:
  \[
  -2A + 3B = B
  \]
  Simplify:
  \[
  -2A + 3B = B \implies -2A = -2B \implies A = B
  \]
  
  Substitute \( A = B \) into \( A + B = 6 \):
  \[
  B + B = 6 \implies 2B = 6 \implies B = 3
  \]
  Since \( A = B \), we have \( A = 3 \).
  
  Therefore, \( A \times B = 3 \times 3 = 9 \).

### 3. \( A(x - 2) + B(x + 1) \equiv 2x + 5 \)
We are asked to calculate the value of \( A \cdot B \).

- Expand both sides:
  \[
  A(x - 2) + B(x + 1) = A(x) - 2A + B(x) + B = (A + B)x + (-2A + B)
  \]
  
  Compare the coefficients of \( x \):
  \[
  A + B = 2
  \]
  
  And compare the constant terms:
  \[
  -2A + B = 5
  \]
  
  Solve the system of equations:
  - From \( A + B = 2 \), we can express \( B \) as \( B = 2 - A \).
  - Substitute \( B = 2 - A \) into \( -2A + B = 5 \):
    \[
    -2A + (2 - A) = 5 \implies -2A + 2 - A = 5 \implies -3A + 2 = 5 \implies -3A = 3 \implies A = -1
    \]
  - Substitute \( A = -1 \) into \( A + B = 2 \):
    \[
    -1 + B = 2 \implies B = 3
    \]
  
  Therefore, \( A \cdot B = -1 \times 3 = -3 \).

### 4. \( P(x) = 7 + 2x^{a+b} + x^{2a-b} + 5x^{3c+d} - x^{4c-d} \)
We are asked to calculate the value of \( a + b + n + m \), assuming the expression is complete and ordered.

- Analyze the exponents in each term:
  \[
  2x^{a+b} + x^{2a-b} + 5x^{3c+d} - x^{4c-d}
  \]
  To be consistent, all powers of \( x \) must align:
  - The powers must align as \( a + b \), \( 2a - b \), \( 3c + d \), and \( 4c - d \).
  However, there are not enough details to uniquely determine \( a \), \( b \), \( c \), or \( d \), so additional information might be necessary for this problem. However, the general approach would involve comparing exponents and solving similar to the above.

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Would you like further clarification on any of these solutions?

Here are some relative questions to extend your understanding:
1. What are the general strategies for solving systems of equations involving multiple variables?
2. How do you approach comparing coefficients in polynomial identities?
3. Can you solve other types of equations where terms involve different powers of \( x \)?
4. How would this system change if one equation were quadratic?
5. What happens if we introduce complex numbers to these types of problems?

**Tip:** When solving equations with multiple variables, always aim to simplify and compare coefficients where possible, as this helps isolate each variable step-by-step.

1. Si 5x + 7 ≡ (m + n)x + 2m – 1; calcula el valor «m – n».
2. Si A(x – 2) + B(x + 3) ≡ 6x + B, determina el valor de «A × B».
3. Si A(x – 2) + B(x + 1) ≡ 2x + 5, calcula el valor de «A × B».
4. Si P(x) = 7 + 2x^{a+b} + x^{2a-b} + 5x^{3c+d} – x^{4c-d}, es completo y ordenado, calcula el valor de «a + b + n + m».

Explore and solve algebraic expressions, including linear systems of equations and polynomial expansions. Learn to calculate variable products and simplify complex expressions with step-by-step solutions. Suitable for Grades 9-11.

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