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Pertanyaan

Akar-akar persamaan 1 − 28 ⋅ 3 − x + 27 ⋅ 9 − x = 0 adalah x 1 ​ dan x 2 ​ . Nilai x 1 ​ + x 2 ​ = ...

Akar-akar persamaan  adalah dan . Nilai  

  1. 1

  2. 3

  3. 4

  4. 5

  5. 28

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N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B

jawaban yang tepat adalah B

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Pembahasan

Pembahasan
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Ingat kembali: Misalkan: Sehingga diperoleh perhitungan: Untuk Untuk Sehingga, Dengan demikian, hasil dari adalah Jadi, jawaban yang tepat adalah B

Ingat kembali:

a to the power of n equals stack stack a cross times a cross times a cross times... cross times a with underbrace below with a space s e b a n y a k space n below comma space a not equal to 0

open parentheses a to the power of m close parentheses to the power of n equals a to the power of m times n end exponent comma space a not equal to 0

1 over a to the power of n equals a to the power of negative n end exponent comma space a not equal to 0

a to the power of f open parentheses x close parentheses end exponent equals a to the power of g open parentheses x close parentheses end exponent rightwards arrow f open parentheses x close parentheses equals g open parentheses x close parentheses comma space a not equal to 0

a to the power of 0 equals 1 comma space a not equal to 0

Misalkan:

3 to the power of negative x end exponent equals y 

Sehingga diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 minus 28 times 3 to the power of negative x end exponent plus 27 times 9 to the power of negative x end exponent end cell equals 0 row cell 1 minus 28 times 3 to the power of negative x end exponent plus 27 times open parentheses 3 squared close parentheses to the power of negative x end exponent end cell equals 0 row cell 1 minus 28 times 3 to the power of negative x end exponent plus 27 times 3 to the power of open parentheses negative x close parentheses times 2 end exponent end cell equals 0 row cell 1 minus 28 times 3 to the power of negative x end exponent plus 27 times open parentheses 3 to the power of negative x end exponent close parentheses squared end cell equals 0 row cell 1 minus 28 a plus 27 a squared end cell equals 0 row cell 27 y squared minus 28 y plus 1 end cell equals cell 0 space open parentheses difaktorkan close parentheses end cell row cell open parentheses 27 y minus 1 close parentheses open parentheses y minus 1 close parentheses end cell equals cell 0 space end cell row blank blank cell y equals 1 over 27 space atau space y equals 1 end cell end table  

Untuk y equals 1 over 27 

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 to the power of negative x end exponent end cell equals y row cell 3 to the power of negative x end exponent end cell equals cell 1 over 27 end cell row cell 3 to the power of negative x end exponent end cell equals cell 1 over 3 cubed end cell row cell 3 to the power of negative x end exponent end cell equals cell 3 to the power of negative 3 end exponent end cell row cell negative x end cell equals cell negative 3 end cell row cell x subscript 1 end cell equals 3 end table  

Untuk y equals 1

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 to the power of negative x end exponent end cell equals y row cell 3 to the power of negative x end exponent end cell equals 1 row cell 3 to the power of negative x end exponent end cell equals cell 3 to the power of 0 end cell row cell negative x end cell equals 0 row cell x subscript 2 end cell equals 0 end table  

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus x subscript 2 space end subscript end cell equals cell 3 plus 0 end cell row blank equals 3 end table 

Dengan demikian, hasil dari x subscript 1 plus x subscript 2 space end subscript adalah  table attributes columnalign right center left columnspacing 0px end attributes row blank blank 3 end table 

Jadi, jawaban yang tepat adalah B

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