Materi : Bentuk Akar dan Pangkat
( x³/y² )⁶ ÷ ( x²/y )³
= x¹⁸/y¹² × y³/x⁶
= x¹²/y⁹
---
( m/n )⁵ × ( m/n )-⁵ = ( m/n )⁰ = 1
( a²/⁵ × a¹/² )/a¹/⁴ = a⁴/¹⁰+⁵/¹⁰-¹/⁴ = a⁹/¹⁰-¹/⁴ = a²⁶/⁴⁰ = a¹³/²⁰
Semoga bisa membantu
[tex] \boxed{ \colorbox{darkblue}{ \sf{ \color{lightblue}{ answered\:by\: BLUEBRAXGEOMETRY}}}} [/tex]
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Materi : Bentuk Akar dan Pangkat
( x³/y² )⁶ ÷ ( x²/y )³
= x¹⁸/y¹² × y³/x⁶
= x¹²/y⁹
---
( m/n )⁵ × ( m/n )-⁵ = ( m/n )⁰ = 1
---
( a²/⁵ × a¹/² )/a¹/⁴ = a⁴/¹⁰+⁵/¹⁰-¹/⁴ = a⁹/¹⁰-¹/⁴ = a²⁶/⁴⁰ = a¹³/²⁰
---
Semoga bisa membantu
[tex] \boxed{ \colorbox{darkblue}{ \sf{ \color{lightblue}{ answered\:by\: BLUEBRAXGEOMETRY}}}} [/tex]