174k views
0 votes
Simplify quantity 18 x minus 6 over 9 x to the fifth power all over quantity 15 x plus 5 over 21 x squared

User Asaka
by
7.9k points

2 Answers

3 votes

Answer:
(98(3x-1))/(3^7.5x^3(3x+1))

Explanation:

Given expression,
(18x-6/(9x)^5)/(15x+5/(21)^2)

=
((18x-6)* 21^2x^2)/(9^5.x^5* (15x+5))

=
((18x-6).7^2)/(x^3.9^4(15x+5))

=
(2* 7^2(3x-1))/(3^7x^3(15x+5))

=
(2* 49(3x-1))/(3^7x^3.5(3x+1))

=
(98(3x-1))/(3^7.5x^3(3x+1))

User Nomesh Gajare
by
8.6k points
3 votes
Going off of your description, I got and solved this equation:


( (18x-6)/(9x^(5) ))/( (15x+5)/(21x^(2) ) )

I simplified it to:


( (6(3x-1))/(9x^(5)) )/( (5(3x+1))/(21 x^(2) ) )

I then simplified again and flipped the second fraction to get a multiplication problem:


(2(3x-1))/(3 x^(5) )( (21 x^(2) )/(5(3x+1))

Which then simplifies to your answer:


(14(3x-1))/(5 x^(3)(3x+1))
User Bobics
by
8.4k points

No related questions found