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Please solve the problem

I want whole process​

Please solve the problem I want whole process​-example-1
User Nguyenkha
by
7.6k points

2 Answers

10 votes

Answer:

1

Explanation:

Solving :


(x^(p(q-r)) )/(x^(q(p-r)) ) ÷
((x^(q) )/(x^(p) ))^(r)


(x^(p(q-r)) )/(x^(q(p-r)) ) ÷
[ ({x^(q-p) })^(r)| or| x^(r(q-p)) ]


{x}^(pq-pr-qp+qr) ÷
x^(rq-rp)


x^(pq-pq+qr-qr+pr-pr)

⇒ x⁰

1

User Afeshia
by
8.9k points
5 votes

Answer:

1

Explanation:

⠀⠀⠀⠀□ Question


\Large \frac{ {x}^(p(q - r)) }{ {x}^(q(p - r)) } / {( \frac{ {x}^(q) }{{x}^(p) } })^(r)

⠀⠀⠀⠀■ Solution

  • Multiplying the power


\sf \Large{ \frac{ {x}^(pq - pr) }{ {x}^(pq - qr) } } / \frac{ {x}^(qr) }{ {x}^(pr) }

  • Changing ÷ sign into × by doing reciprocal


\sf\Large{ \frac{ {x}^(pq - pr) }{ {x}^(pq - qr) } } * \frac{ {x}^(pr) }{ {x}^(qr) }

  • As the base(x) is same, if the value is multipying so the power will add, if the value is dividing so the power will subtract.


\sf {x}^(pq - pr - pq + qr) * {x}^(pr - qr) \\ \\ \sf {x}^(qr - pr) * {x}^(pr - qr) \\ \\ \sf {x}^(qr - qr + pr - pr)

  • If the power of any value become 0,so it's value is always equal to 1


\sf \Large{ {x}^(0) } \\ \\ \boxed {\large {\rightarrow1 }}

User Tim Sylvester
by
8.1k points

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