Math Problem: Simplifying The Expression Step-by-Step

Hey guys, let's dive into this cool math problem! We're gonna break down how to simplify the expression: ((1 / (a − √a c)) + (1 / (a + √a c))) / (1 − c/a)⁻¹. Don't worry, it looks a bit intimidating at first, but trust me, we'll get through it together. Our goal here is to transform this complex expression into something much simpler and easier to understand. This process involves a few key algebraic techniques that are super useful for any math enthusiast. So, let's roll up our sleeves and get started! We will use the main keywords in the beginning of each paragraph, use bold, italic and strong tags.

First, let's address the elephant in the room: the fractions. We have two fractions to start with, and a fraction within a fraction. The initial step in tackling this problem involves simplifying the expression inside the parentheses. We're going to use a common denominator to add these two fractions. This means we're going to multiply the first fraction's numerator and denominator by (a + √ac) and the second fraction's numerator and denominator by (a - √ac). This gives us: ((a + √ac) / ((a − √ac)(a + √ac))) + ((a - √ac) / ((a + √ac)(a - √ac))).

Next up, simplify the denominators. Remember that (a - b)(a + b) equals a² - b². Applying this to our problem, the denominators become a² - (√ac)², which simplifies to a² - ac. Now, our expression looks like this: ((a + √ac) + (a - √ac)) / (a² - ac). Combining the numerators, we notice that the +√ac and -√ac cancel each other out. This leaves us with (2a) / (a² - ac). Pretty neat, huh?

Now, let's deal with the second part of the equation: (1 - c/a)⁻¹. This is where things get even more interesting. The negative exponent means we need to take the reciprocal of the term. First, simplify the expression inside the parentheses. To do this, find a common denominator, which in this case is 'a'. So, 1 becomes a/a. Therefore, (1 - c/a) becomes (a/a - c/a), which equals (a - c) / a. Now, we take the reciprocal because of the negative exponent, flipping the fraction to become a / (a - c). We're making great progress, aren't we?

Finally, let's combine everything! Our initial expression was ((1 / (a − √ac)) + (1 / (a + √ac))) / (1 − c/a)⁻¹. We've simplified the first part to (2a) / (a² - ac) and the second part to a / (a - c). Now, we divide the first simplified fraction by the second simplified fraction. Remember, dividing by a fraction is the same as multiplying by its reciprocal. So, we multiply (2a) / (a² - ac) by (a - c) / a.

This gives us: ((2a) / (a² - ac)) * (a / (a - c)). We can simplify this further by factoring 'a' out of the denominator in the first fraction. This gives us (2a) / (a(a - c)) * (a - c) / a. Notice that we can cancel out the 'a' in the numerator and denominator, as well as the (a - c) terms. This leaves us with the final simplified answer. The ability to manipulate and simplify such expressions is fundamental to higher-level mathematics. So, keeping up with these steps is an important exercise. By working through each step carefully, you not only solve the problem, but also reinforce your understanding of core algebraic principles. Keep it up!

Deep Dive into the Simplification Process

Now that we've crunched the numbers and reached our final answer, let's rewind a bit and really understand the why behind each step. It's one thing to follow a set of instructions, but quite another to truly grasp the underlying concepts. This will help you become a master of simplification.

Let's revisit the initial step: adding the fractions. The concept of finding a common denominator is at the heart of this. When we add or subtract fractions, we need to ensure they have the same denominator. This is like comparing apples to apples. If we try to add 1/2 and 1/3 directly, it's not straightforward. But if we convert them to 3/6 and 2/6, we can easily see that the sum is 5/6. The common denominator allows us to combine the numerators accurately. This concept is so fundamental that you'll use it again and again as you tackle more complex math problems. It's the foundation upon which more advanced techniques are built.

Next, we used the difference of squares: (a - b)(a + b) = a² - b². This is a classic algebraic identity that can save you a ton of time and effort. Instead of multiplying out the terms individually, recognizing this pattern allows for a quick and elegant simplification. Whenever you see a product of two binomials that look like (x - y) and (x + y), immediately think of the difference of squares. The ability to recognize and apply such identities is a hallmark of a strong mathematician. This is why practicing these patterns is so important for improving your math skills. Being able to spot these patterns can make a huge difference in your ability to solve problems quickly and efficiently.

Then, let's talk about negative exponents. A negative exponent is simply a shorthand way of saying