Given:
[tex]\frac{x_1+x_2}{9}\div\frac{ea^4x_1^2-ea^4x_2^2}{81ua}[/tex]
Required:
We need to simplify the given expression.
Explanation:
[tex]Use\text{ }\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}.[/tex]
[tex]\frac{x_1+x_2}{9}\div\frac{ea^4x_1^2-ea^4x_2^2}{81ua}=\frac{x_1+x_2}{9}\times\frac{81ua}{ea^4x_1^2-ea^4x_2^2}[/tex]
[tex]=\frac{x_1+x_2}{9}\times\frac{81ua}{ea^4x_1^2-ea^4x_2^2}[/tex]
Take the common term out.
[tex]=\frac{x_1+x_2}{9}\times\frac{81ua}{ea^4(x_1^2-x_2^2)}[/tex][tex]Use\text{ }a^2-b^2=(a-b)(a+b).[/tex]
[tex]=\frac{x_1+x_2}{9}\times\frac{81ua}{ea^4(x_1-x_2)(x_1+x_2)}[/tex][tex]Ca\text{ncel out the common term }(x_1+x_2).[/tex]
[tex]=\frac{1}{9}\times\frac{81ua}{ea^4(x_1-x_2)}[/tex]
Cancel out the term a.
[tex]=\frac{1}{9}\times\frac{81u}{ea^3(x_1-x_2)}[/tex]
Cancel out 9 multiples.
[tex]=\frac{9u}{ea^3(x_1-x_2)}[/tex]
Final answer:
[tex]\frac{x_1+x_2}{9}\div\frac{ea^4x_1^2-ea^4x_2^2}{81ua}=\frac{9u}{ea^3(x_1-x_2)}[/tex]
