e.) $(a+b)^2 \times(a+b)^{-2}$
Answer (1)
Rewrite the expression using the property of negative exponents: ( a + b ) − 2 = ( a + b ) 2 1 .
Multiply the terms: ( a + b ) 2 × ( a + b ) 2 1 = ( a + b ) 2 ( a + b ) 2 .
Simplify the fraction: If a + b = 0 , then ( a + b ) 2 ( a + b ) 2 = 1 .
If a + b = 0 , the expression is undefined. Therefore, the simplified expression is 1 .
Explanation
Understanding the Problem We are asked to simplify the expression ( a + b ) 2 " , ( a + b ) − 2 . This expression involves exponents, and we can use the properties of exponents to simplify it.
Rewriting the Expression We can rewrite the expression using the property x − n = x n 1 . So, ( a + b ) − 2 = ( a + b ) 2 1 . Therefore, the expression becomes: ( a + b ) 2 " , ( a + b ) − 2 = ( a + b ) 2 " , ( a + b ) 2 1
Simplifying the Expression Now we can simplify the expression by multiplying the terms: ( a + b ) 2 " , ( a + b ) 2 1 = ( a + b ) 2 ( a + b ) 2
Final Simplification If a + b = 0 , then any non-zero number divided by itself is 1. Therefore, ( a + b ) 2 ( a + b ) 2 = 1 If a + b = 0 , then the expression is undefined because we would be dividing by zero.
Examples
In electrical engineering, when analyzing circuits, you might encounter expressions involving impedance. Simplifying these expressions, similar to the one above, helps in determining the overall behavior of the circuit. For instance, if ( a + b ) represents a complex impedance, the given expression could arise in calculations related to power transfer or resonance conditions. Simplifying such expressions allows engineers to quickly assess the circuit's performance and make necessary adjustments. Understanding how to manipulate exponents and algebraic expressions is crucial for efficient circuit analysis.
