Respuesta :
Answer:
[tex]\dfrac{4(3x^2y^4)^3}{(2x^3y^5)^4}=\dfrac{27}{4x^6y^8}[/tex]
Step-by-step explanation:
Given:
The expression to simplify is given as:
[tex]\dfrac{4(3x^2y^4)^3}{(2x^3y^5)^4}[/tex]
First, we will simplify the numerator and denominator separately using the law of indices:
[tex](ka^mb^n)^p=k^pa^{m\times p}b^{n\times p}[/tex]
The numerator is simplified as:
[tex]4(3x^2y^4)^3=4(3^3x^{2\times 3}y^{4\times 3})=4(27x^6y^{12})[/tex]
The denominator is simplified as:
[tex](2x^3y^5)^4=2^4x^{3\times 4}y^{5\times 4}=16x^{12}y^{20}[/tex]
Now, we divide the simplified numerator by the simplified denominator. This gives,
[tex]=\frac{4(27x^6y^{12})}{16x^{12}y^{20}}\\=\frac{4\times27}{16}\times \frac{x^6}{x^{12}}\times \frac{y^{12}}{y^{20}}[/tex]
Now, we simplify using another law of indices which is given as:
[tex]\frac{a^m}{a^n}=a^{m-n}[/tex]
[tex]=\frac{4\times27}{16}\times \frac{x^6}{x^{12}}\times \frac{y^{12}}{y^{20}}\\=\frac{27}{4}\times x^{6-12}\times y^{12-20}\\=\frac{27}{4}x^{-6}y^{-8}[/tex]
Now, we write the answer using only positive exponents and thus we use the given law of indices:
[tex]a^{-m}=\frac{1}{a^m}[/tex]
Therefore, the simplified form is:
[tex]\dfrac{27}{4}x^{-6}y^{-8}=\dfrac{27}{4x^6y^8}[/tex]
