Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Answer:
1
Step-by-step explanation:
You want to simplify the given complex fraction involving fractional powers.
Simplify
We can define a = x^(1/2) and b = y^(1/2) to remove fractional powers from the expression. The rest of it involves factoring the sum of cubes and the difference of squares. Common factors are cancelled.
[tex]\left(\dfrac{a^3+b^3}{a^2-b^2}-\dfrac{a^2}{a+b}-\dfrac{b^2}{a-b}\right)\div\dfrac{ab}{a+b}\\\\=\left(\dfrac{a^3+b^3}{a^2-b^2}-\dfrac{a^2}{a+b}-\dfrac{b^2}{a-b}\right)\times\dfrac{a+b}{ab}\\\\=\left(\dfrac{(a^3+b^3)-a^2(a-b)-b^2(a+b)}{a^2-b^2}\right)\times\dfrac{a+b}{ab}\\\\=\dfrac{(a^3+b^3-a^3+a^2b-ab^2-b^3)(a+b)}{ab(a-b)(a+b)}=\dfrac{a^2b-ab^2}{ab(a-b)}=\dfrac{a^2b-ab^2}{a^2b-ab^2}\\\\=\boxed{1}[/tex]
Note that there are no instances of 'a' or 'b' left in the expression, so we do not need to back-substitute for 'a' and/or 'b' in the simplified form.
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.