Question

Q. Chiiiii`
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b: 25
c: 1/9
[tex] \sf {c}^{ - 1} - 2 {b}^{ \frac{1}{2} } = ...[/tex]
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Answer 1

Materi : Bentuk Akar dan Pangkat

[tex][tex] \sf {c}^{ - 1} - 2 {b}^{ \frac{1}{2} } = ...[/tex][/tex]

{ b = 25 , c = ⅑ }

c-¹ - 2b¹/²

= (⅑)-¹ - 2(25)¹/²

= (9-¹)-¹ - 2(5²)¹/²

= 9¹ - 2(5¹)

= 9 - 2(5)

= 9 - 10

= -1

Semoga bisa membantu

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

Answer 2

Jawaban:

Penyelesaian :

Diketahui :

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

Jawaban :

[tex] \sf {c}^{ - 1} - {2b}^{ \frac{1}{2} } [/tex]

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

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

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

[tex] \sf = {9}^{( - 1 \times - 1)} - 2( {5}^{1} )[/tex]

[tex] \sf = {9}^{(1 \times 1)} - 2(5)[/tex]

[tex] \sf = {9}^{1} - (10)[/tex]

[tex] \sf = 9 - 10[/tex]

[tex] \sf = \red{ - 1}[/tex]

'비상' (Svt)