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Nilai dari persamaan  adalah...

Pertanyaan

Nilai dari persamaan open parentheses 1 over 9 close parentheses to the power of 3 minus 3 x end exponent less or equal than open parentheses 1 over 27 close parentheses to the power of x minus x squared end exponent adalah...

  1. x less or equal than negative 2 space atau space x greater or equal than negative 1 

  2. x less or equal than negative 2 space atau space x greater or equal than 1 

  3. x less or equal than 1 space atau space x greater or equal than 2 

  4. 1 less or equal than x less or equal than 2 

  5. negative 2 less or equal than x less or equal than negative 1  

Pembahasan Soal:

Pada pertidaksamaan eksponen berlaku :

a to the power of f open parentheses x close parentheses end exponent less or equal than a to the power of g open parentheses x close parentheses end exponent rightwards arrow f open parentheses x close parentheses less or equal than g open parentheses x close parentheses 

Sehingga untuk mencari nilai dari persamaan open parentheses 1 over 9 close parentheses to the power of 3 minus 3 x end exponent less or equal than open parentheses 1 over 27 close parentheses to the power of x minus x squared end exponent dapat kita uraikan sebagai berikut :

table attributes columnalign right center left columnspacing 2px end attributes row cell open parentheses 1 over 9 close parentheses to the power of 3 minus 3 x end exponent end cell less or equal than cell open parentheses 1 over 27 close parentheses to the power of x minus x squared end exponent end cell row cell open parentheses 1 squared over 3 squared close parentheses to the power of 3 minus 3 x end exponent end cell less or equal than cell open parentheses 1 cubed over 3 cubed close parentheses to the power of x minus x squared end exponent end cell row cell open parentheses 1 third close parentheses to the power of 2 open parentheses 3 minus 3 x close parentheses end exponent end cell less or equal than cell open parentheses 1 third close parentheses to the power of 3 open parentheses x minus x squared close parentheses end exponent end cell row cell open parentheses 1 third close parentheses to the power of 6 minus 6 x end exponent end cell less or equal than cell open parentheses 1 third close parentheses to the power of 3 x minus 3 x squared end exponent end cell row cell 6 minus 6 x end cell less or equal than cell 3 x minus 3 x squared end cell row cell 3 x squared minus 6 x minus 3 x plus 6 end cell less or equal than 0 row cell 3 x squared minus 9 x plus 6 end cell less or equal than 0 end table 

Kemudian kita faktorkan pembuat nol nya :

table attributes columnalign right center left columnspacing 2px end attributes row cell 3 x squared minus 9 x plus 6 end cell equals 0 row cell x squared minus 3 x plus 2 end cell equals 0 row cell open parentheses x minus 2 close parentheses open parentheses x minus 1 close parentheses end cell equals 0 row x equals 2 row atau blank blank row x equals 1 end table 

Selanjutnya nilai x equals 1 space dan x equals 2  kita letakkan pada  garis bilangan dan selanjutnya kita buat titik uji x equals 0 untuk mengetahui daerah positif dan negatifnya.

f open parentheses x close parentheses equals 3 x squared minus 9 x plus 6 f open parentheses 0 close parentheses equals 3 open parentheses 0 close parentheses squared minus 9 open parentheses 0 close parentheses plus 6 equals 6 

Karena didapatkan hasilnya positif maka daerah garis bilangannya sebagai berikut :

 

Nah, karena pertidaksamaannya tadi f open parentheses x close parentheses less or equal than 0 sehingga yang di arsir adalah daerah negatif

 

Jadi, Nilai dari persamaan open parentheses 1 over 9 close parentheses to the power of 3 minus 3 x end exponent less or equal than open parentheses 1 over 27 close parentheses to the power of x minus x squared end exponent adalah 1 less or equal than x less or equal than 2

Oleh karena itu, jawaban yang tepat adalah D

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 11 Juli 2021

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Pertanyaan yang serupa

Himpunan penyelesaian dari adalah...

Pembahasan Soal:

Ingat bahwa pentuk pertidaksamaan eksponen:

a to the power of f open parentheses x close parentheses end exponent less or equal than a to the power of g open parentheses x close parentheses end exponent rightwards arrow f open parentheses x close parentheses less or equal than g open parentheses x close parentheses 

Sehingga diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 1 over 8 close parentheses to the power of 8 plus 2 x minus x squared end exponent end cell less or equal than cell open parentheses 1 over 16 close parentheses to the power of x plus 2 end exponent end cell row cell open parentheses 2 to the power of negative 3 end exponent close parentheses to the power of 8 plus 2 x minus x squared end exponent end cell less or equal than cell open parentheses 2 to the power of negative 4 end exponent close parentheses to the power of x plus 2 end exponent end cell row cell 2 to the power of negative 3 open parentheses 8 plus 2 x minus x squared close parentheses end exponent end cell less or equal than cell 2 to the power of negative 4 open parentheses x plus 2 close parentheses end exponent end cell row cell 2 to the power of negative 24 minus 6 x plus 3 x squared end exponent end cell less or equal than cell 2 to the power of negative 4 x minus 8 end exponent end cell row blank blank blank row blank rightwards double arrow cell negative 24 minus 6 x plus 3 x squared less or equal than negative 4 x minus 8 end cell row cell negative 24 minus 6 x plus 3 x squared plus 4 x plus 8 end cell less or equal than 0 row cell 3 x squared minus 2 x minus 16 end cell less or equal than 0 row cell open parentheses 3 x minus 8 close parentheses open parentheses x plus 2 close parentheses end cell less or equal than 0 row x equals cell 8 over 3 space atau space straight x equals negative 2 end cell row blank blank blank end table 

titik uji x equals 0

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x squared minus 2 x minus 16 end cell less or equal than 0 row cell open parentheses 0 close parentheses minus open parentheses 0 close parentheses minus 16 end cell less or equal than 0 row cell negative 16 end cell less or equal than cell 0 space open parentheses b e n a r close parentheses end cell end table

Sehingga himpunanan penyelesaiannya open curly brackets x vertical line minus 2 less or equal than x less or equal than 8 over 3 close curly brackets

Jadi, tidak ada jawaban yang tepat

1

Roboguru

Bentuk sederhana dari adalah ....

Pembahasan Soal:

begin mathsize 14px style open parentheses fraction numerator 4 a to the power of negative 2 end exponent b squared c over denominator 12 a to the power of negative 5 end exponent b to the power of 4 c to the power of negative 1 end exponent end fraction close parentheses to the power of negative 1 end exponent equals open parentheses fraction numerator a to the power of negative 2 minus left parenthesis negative 5 right parenthesis end exponent b to the power of 2 minus 4 end exponent c to the power of 1 minus open parentheses negative 1 close parentheses end exponent over denominator 3 end fraction close parentheses to the power of negative 1 end exponent  equals open parentheses fraction numerator a cubed b to the power of negative 2 end exponent c squared over denominator 3 end fraction close parentheses to the power of negative 1 end exponent  equals fraction numerator a to the power of negative 3 end exponent b squared c to the power of negative 2 end exponent over denominator 3 to the power of negative 1 end exponent end fraction  equals fraction numerator 3 b squared over denominator a cubed c squared end fraction end style

0

Roboguru

Nilai  yang memenuhi  untuk semua  adalah ....

Pembahasan Soal:

Diketahui undefined, maka:

begin mathsize 14px style open parentheses 0 comma 144 close parentheses to the power of negative x squared plus 3 x minus 8 end exponent less than open parentheses 0 comma 12 close parentheses to the power of x squared minus 2 x plus a end exponent open parentheses open parentheses 0 comma 144 close parentheses squared close parentheses to the power of negative x squared plus 3 x minus 8 end exponent less than open parentheses 0 comma 12 close parentheses to the power of x squared minus 2 x plus a end exponent open parentheses 0 comma 144 close parentheses to the power of negative 2 x squared plus 6 x minus 16 end exponent less than open parentheses 0 comma 12 close parentheses to the power of x squared minus 2 x plus a end exponent end style  

karena basisnya bernilai diantara begin mathsize 14px style 0 end style dan begin mathsize 14px style 1 end style, maka untuk pangkat tandanya dibalik menjadi:

begin mathsize 14px style negative 2 x squared plus 6 x minus 16 greater than x squared minus 2 x plus a minus 3 x squared plus 8 x minus 16 minus a greater than 0 end style 

solusi dari pertidaksamaan di atas bisa dicari dengan cara:

begin mathsize 14px style x subscript 1 comma 2 end subscript equals fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction x subscript 1 comma 2 end subscript equals fraction numerator negative 8 plus-or-minus square root of 8 squared minus 4 open parentheses negative 3 close parentheses open parentheses negative 16 minus a close parentheses end root over denominator 2 open parentheses negative 3 close parentheses end fraction x subscript 1 comma 2 end subscript equals fraction numerator negative 8 plus-or-minus square root of 64 plus 12 open parentheses negative 16 minus a close parentheses end root over denominator negative 6 end fraction x subscript 1 comma 2 end subscript equals fraction numerator negative 8 plus-or-minus square root of 64 minus 192 minus 12 a end root over denominator negative 6 end fraction x subscript 1 comma 2 end subscript equals fraction numerator negative 8 plus-or-minus square root of negative 128 minus 12 a end root over denominator negative 6 end fraction end style  

agar persamaan di atas memiliki solusi, maka:

begin mathsize 14px style square root of negative 128 minus 12 a end root greater than 0 space space space space minus 128 minus 12 a greater than 0 space space space space space space space space space space space space space minus 12 a greater than 128 space space space space space space space space space space space space space space space space space space space space a greater than fraction numerator 128 over denominator negative 12 end fraction space space space space space space space space space space space space space space space space space space space space a greater than negative 10 8 over 12 space space space space space space space space space space space space space space space space space space space space a greater than negative 10 2 over 3 end style 

Jadi, jawaban yang benar adalah B.

0

Roboguru

Jika , maka  mempunyai penyelesaian ...

Pembahasan Soal:

Untuk mencari himpunan penyelesaian pertidaksamaan fraction numerator 3 plus 3 a to the power of x over denominator 1 plus a to the power of x end fraction less than a to the power of x, pertama kita misalkan u equals a to the power of x.


table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 3 plus 3 a to the power of x over denominator 1 plus a to the power of x end fraction end cell less than cell a to the power of x end cell row cell fraction numerator 3 plus 3 u over denominator 1 plus u end fraction end cell less than u row cell fraction numerator 3 plus 3 u over denominator 1 plus u end fraction minus u end cell less than 0 row cell fraction numerator open parentheses 3 plus 3 u close parentheses minus u open parentheses 1 plus u close parentheses over denominator 1 plus u end fraction end cell less than 0 row cell fraction numerator 3 plus 3 u minus u minus u squared over denominator 1 plus u end fraction end cell less than 0 row cell fraction numerator negative u squared plus 2 u plus 3 over denominator 1 plus u end fraction end cell less than 0 end table


Cari pembuat 0 dari penyebut dan pembilangnya.


table attributes columnalign right center left columnspacing 0px end attributes row cell negative u squared plus 2 u plus 3 end cell equals 0 row cell open parentheses negative u plus 3 close parentheses open parentheses u plus 1 close parentheses end cell equals 0 row u equals cell 3 space atau space u equals negative 1 end cell end table


table attributes columnalign right center left columnspacing 0px end attributes row cell 1 plus u end cell equals 0 row u equals cell negative 1 end cell end table


Setelah itu, buat garis bilangan dan uji untuk mendapatkan tanda dari garis bilangan tersebut.



Maka kita mendapatkan 2 daerah penyelesaian, yaitu u less than negative 1 atau u greater than 3. Kita kembalikan u equals a to the power of x.

  1. a to the power of x less than negative 1 tidak mempunyai penyelesaian karena untuk 0 less than a less than 1 maka a to the power of x greater than 0.
  2. Untuk 0 less than a less than 1, maka a to the power of x greater than 3 mempunyai penyelesaian x less than log subscript a 3.

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

Penyelesaian dari : adalah ...

Pembahasan Soal:

Karena kita ingin menyelesaikan bentuk pertidaksamaan eksponen, maka hal yang perlu kamu perhatikan lebih dulu adalah nilai basisnya, apakah bernilai lebih dari 1 atau antara 0 sampai 1. Jika kita uraikan soalnya terlebih dahulu, maka diperoleh nilai basisnya, yaitu 1 half. Sehingga, tanda pertidaksamaannya tetap. Penjelasan lebih lengkapnya bisa kamu lihat di bawah ini

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 1 half close parentheses to the power of straight x squared minus straight x minus 12 end exponent end cell less or equal than cell open parentheses 1 half close parentheses to the power of straight x minus 4 end exponent end cell row cell straight x squared minus straight x minus 12 end cell less or equal than cell straight x minus 4 end cell row cell straight x squared minus straight x minus straight x end cell less or equal than cell negative 4 plus 12 end cell row cell straight x squared minus 2 straight x end cell less or equal than 8 row cell straight x squared minus 2 straight x minus 8 end cell less or equal than 0 row cell open parentheses straight x minus 4 close parentheses open parentheses straight x plus 2 close parentheses end cell less or equal than 0 row blank blank blank end table

Dari persamaan di atas, maka nilai x adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell straight x minus 4 end cell less or equal than 0 row straight x less or equal than cell 4 space atau space end cell row cell straight x plus 2 end cell less or equal than 0 row straight x less or equal than cell negative 2 end cell end table

Jika dibuat ke dalam garis bilangan 

Jadi, himpunan penyelesainnya adalah open curly brackets xI minus 2 less or equal than straight x less or equal than 4 comma straight x element of straight R close curly brackets.

0

Roboguru

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