Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
To solve the expression [tex]\(\left(-2 \sqrt[3]{x^3 y}\right)\left(-5 \sqrt[4]{x y^2}\right)\left(3 \sqrt[12]{x^3 y^2}\right)\)[/tex], let's break it down step-by-step:
1. Define each term in the expression:
- First term: [tex]\(-2 \sqrt[3]{x^3 y}\)[/tex]
- Second term: [tex]\(-5 \sqrt[4]{x y^2}\)[/tex]
- Third term: [tex]\(3 \sqrt[12]{x^3 y^2}\)[/tex]
2. Rewrite each term with fractional exponents:
- [tex]\( \sqrt[3]{x^3 y} \)[/tex] is equivalent to [tex]\((x^3 y)^{\frac{1}{3}}\)[/tex]
- [tex]\( \sqrt[4]{x y^2} \)[/tex] is equivalent to [tex]\((x y^2)^{\frac{1}{4}}\)[/tex]
- [tex]\( \sqrt[12]{x^3 y^2} \)[/tex] is equivalent to [tex]\((x^3 y^2)^{\frac{1}{12}}\)[/tex]
Now the terms are:
- First term: [tex]\(-2 (x^3 y)^{\frac{1}{3}}\)[/tex]
- Second term: [tex]\(-5 (x y^2)^{\frac{1}{4}}\)[/tex]
- Third term: [tex]\( 3 (x^3 y^2)^{\frac{1}{12}} \)[/tex]
3. Combine the terms by multiplication:
[tex]\[ \left(-2 (x^3 y)^{\frac{1}{3}}\right) \left(-5 (x y^2)^{\frac{1}{4}}\right) \left(3 (x^3 y^2)^{\frac{1}{12}}\right) \][/tex]
4. Multiply the constants:
Combine [tex]\(-2\)[/tex], [tex]\(-5\)[/tex], and [tex]\(3\)[/tex]:
[tex]\[ (-2) \times (-5) \times 3 = 30 \][/tex]
5. Combine the exponent terms:
When multiplying terms with the same base, you add the exponents. So you need to combine the exponents for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
- For [tex]\(x\)[/tex]:
[tex]\[ x^{\frac{3}{3}} \times x^{\frac{1}{4}} \times x^{\frac{3}{12}} \][/tex]
Rewrite the exponents with a common denominator (which is 12):
[tex]\[ x^1 \times x^{\frac{3}{12}} \times x^{\frac{1}{4}} = x^{\frac{12}{12}} \times x^{\frac{1}{4}} = x^{1 + \frac{3}{12} + \frac{1}{4}} = x^{1 + \frac{3}{12} + \frac{3}{12}} = x^{1 + \frac{6}{12}} = x^{1.5} \][/tex]
- For [tex]\(y\)[/tex]:
[tex]\[ y^{\frac{1}{3}} \times y^{\frac{2}{4}} \times y^{\frac{2}{12}} \][/tex]
Rewrite the exponents with a common denominator (which is 12):
[tex]\[ y^{\frac{4}{12}} \times y^{\frac{6}{12}} \times y^{\frac{2}{12}} = y^{\frac{4 + 6 + 2}{12}} = y^{\frac{12}{12}} = y^1 \][/tex]
6. Combine the results:
Now put everything together:
[tex]\[ 30 x^{1.5} y \][/tex]
7. Simplify the final result
It can be rewritten as:
[tex]\[ 30 x^{3/2} y \][/tex]
So, the simplified form of the expression [tex]\(\left(-2 \sqrt[3]{x^3 y}\right)\left(-5 \sqrt[4]{x y^2}\right)\left(3 \sqrt[12]{x^3 y^2}\right)\)[/tex] is [tex]\(30 x^{3/2} y\)[/tex].
1. Define each term in the expression:
- First term: [tex]\(-2 \sqrt[3]{x^3 y}\)[/tex]
- Second term: [tex]\(-5 \sqrt[4]{x y^2}\)[/tex]
- Third term: [tex]\(3 \sqrt[12]{x^3 y^2}\)[/tex]
2. Rewrite each term with fractional exponents:
- [tex]\( \sqrt[3]{x^3 y} \)[/tex] is equivalent to [tex]\((x^3 y)^{\frac{1}{3}}\)[/tex]
- [tex]\( \sqrt[4]{x y^2} \)[/tex] is equivalent to [tex]\((x y^2)^{\frac{1}{4}}\)[/tex]
- [tex]\( \sqrt[12]{x^3 y^2} \)[/tex] is equivalent to [tex]\((x^3 y^2)^{\frac{1}{12}}\)[/tex]
Now the terms are:
- First term: [tex]\(-2 (x^3 y)^{\frac{1}{3}}\)[/tex]
- Second term: [tex]\(-5 (x y^2)^{\frac{1}{4}}\)[/tex]
- Third term: [tex]\( 3 (x^3 y^2)^{\frac{1}{12}} \)[/tex]
3. Combine the terms by multiplication:
[tex]\[ \left(-2 (x^3 y)^{\frac{1}{3}}\right) \left(-5 (x y^2)^{\frac{1}{4}}\right) \left(3 (x^3 y^2)^{\frac{1}{12}}\right) \][/tex]
4. Multiply the constants:
Combine [tex]\(-2\)[/tex], [tex]\(-5\)[/tex], and [tex]\(3\)[/tex]:
[tex]\[ (-2) \times (-5) \times 3 = 30 \][/tex]
5. Combine the exponent terms:
When multiplying terms with the same base, you add the exponents. So you need to combine the exponents for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
- For [tex]\(x\)[/tex]:
[tex]\[ x^{\frac{3}{3}} \times x^{\frac{1}{4}} \times x^{\frac{3}{12}} \][/tex]
Rewrite the exponents with a common denominator (which is 12):
[tex]\[ x^1 \times x^{\frac{3}{12}} \times x^{\frac{1}{4}} = x^{\frac{12}{12}} \times x^{\frac{1}{4}} = x^{1 + \frac{3}{12} + \frac{1}{4}} = x^{1 + \frac{3}{12} + \frac{3}{12}} = x^{1 + \frac{6}{12}} = x^{1.5} \][/tex]
- For [tex]\(y\)[/tex]:
[tex]\[ y^{\frac{1}{3}} \times y^{\frac{2}{4}} \times y^{\frac{2}{12}} \][/tex]
Rewrite the exponents with a common denominator (which is 12):
[tex]\[ y^{\frac{4}{12}} \times y^{\frac{6}{12}} \times y^{\frac{2}{12}} = y^{\frac{4 + 6 + 2}{12}} = y^{\frac{12}{12}} = y^1 \][/tex]
6. Combine the results:
Now put everything together:
[tex]\[ 30 x^{1.5} y \][/tex]
7. Simplify the final result
It can be rewritten as:
[tex]\[ 30 x^{3/2} y \][/tex]
So, the simplified form of the expression [tex]\(\left(-2 \sqrt[3]{x^3 y}\right)\left(-5 \sqrt[4]{x y^2}\right)\left(3 \sqrt[12]{x^3 y^2}\right)\)[/tex] is [tex]\(30 x^{3/2} y\)[/tex].
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
