Question

Q. Chiiiii`
jIka:
a: 8
b: 25
c: 1/9
[tex] \sf \frac{ {3a}^{ \frac{1}{3} }bc }{ {12a}^{ - 2} {bc}^{ - 3} } = ....[/tex]
​

Answer 1

Materi : Bentuk Akar dan Pangkat

[tex][tex] \sf \frac{ {3a}^{ \frac{1}{3} }bc }{ {12a}^{ - 2} {bc}^{ - 3} } = ....[/tex][/tex]

{ a = 8, b = 25, c = ⅑ }

_________________/

[ 3(8)¹/³(25)(⅑) ]/[ 12(8)-²(25)(⅑)-³ ]

= [ 3(2)(5²)(3-²) ]/[ 2².3(2-⁶)(5²)(9³) ]

= [ 2.3-¹.5² ]/[ 2-⁴.3¹⁰.5² ]

= 2⁵/3¹¹

= 32/177.147

Semoga bisa membantu

[tex] \boxed{ \colorbox{darkblue}{ \sf{ \color{lightblue}{ answered\:by\: BLUEBRAXGEOMETRY}}}} [/tex]

Answer 2

Verified answer

Jawaban:

Penyelesaian :

Diketahui :

• a = 8
• b = 25
• c = [tex] \sf \frac{1}{9}[/tex]

Jawaban :

[tex]\sf \frac{ {3a}^{ \frac{1}{3} }bc }{ {12a}^{ - 2} {bc}^{ - 3} }[/tex]

[tex] \sf = \frac{3( \red{8}) {}^{ \frac{1}{3} } (\red{25})( \red{ \frac{1}{9}}) }{12( \red{8}) {}^{ - 2}( \red{25})( \red{ \frac{1}{9}} ) {}^{ - 3} } [/tex]

[tex] \sf = \frac{3 (\red{ {2}^{3} }) {}^{ \frac{1}{3} }( \red{ {5}^{2} }) \frac{1}{9} }{12 (\red{ {2}^{3} }) {}^{ - 2}( \red{ {5}^{2} })( \red{ {9}^{ - 1}} ) {}^{3} } [/tex]

[tex] \sf = \frac{3( {2}^{ \cancel{3}}) {}^{ \frac{1}{ \cancel{3}} } ( {5}^{2} )( \red{{9}^{ - 1}} )}{( \red{ {2}^{2} .3})( {2}^{ - 6} )( {5}^{2} )( {9}^{ - 3} )} [/tex]

[tex] \sf = \frac{3( {2}^{1} ) {5}^{2} ( (\red{ {3}^{2}}) {}^{ - 1}) }{ {2}^{2} . \: 3( {2}^{ - 6} ) {5}^{2} (( \red{ {3}^{2} }) {}^{3} )}[/tex]

[tex] \sf = \frac{3(2) {5}^{2} {3}^{ - 2} }{ {2}^{2}.3( {2}^{ - 6} ) {5}^{2} {3}^{6} } [/tex]

[tex] \sf = \frac{2( {3}^{ - 1}) {5}^{2} }{ {2}^{ - 4} {3}^{7} {5}^{2} } [/tex]

[tex] \sf = {2}^{1 - ( - 4)} {3}^{( -1) - ( - 7)} {5}^{(2 - 2)} [/tex]

[tex] \sf = {2}^{5} {3}^{ - 8} {5}^{0} [/tex]

[tex] \sf = \red{ \frac{ {2}^{5} 1}{ {3}^{8} } }[/tex]

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