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Himpunan penyelesaian dari pertidaksamaan begin mathsize 14px style open vertical bar open vertical bar 3 x plus 6 close vertical bar minus open vertical bar x plus 2 close vertical bar close vertical bar less than open vertical bar x minus 1 close vertical bar end style adalah ...

  1. begin mathsize 14px style open curly brackets x vertical line blank 2 less than x less than 5 comma blank x element of R close curly brackets end style 

  2. begin mathsize 14px style open curly brackets x vertical line minus 5 less than x less than negative 1 comma blank x element of R close curly brackets end style 

  3. begin mathsize 14px style open curly brackets x vertical line x less than negative 1 blank atau blank x greater than 5 comma blank x element of R close curly brackets end style 

  4. begin mathsize 14px style open curly brackets x vertical line minus 1 less than x less than 0 blank atau blank x greater than 2 comma blank x element of R close curly brackets end style 

  5. begin mathsize 14px style open curly brackets x vertical line minus 5 less or equal than x less than 0 blank atau blank x greater than 1 comma blank x element of R close curly brackets end style 

D. Entry

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Dengan definisi nilai mutlak diperoleh,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 3 x plus 6 close vertical bar end cell equals cell open curly brackets table row cell open parentheses 3 x plus 6 close parentheses comma blank open parentheses x plus 2 close parentheses greater or equal than 0 blank a t a u blank x greater or equal than negative 2 end cell row cell negative open parentheses 3 x plus 6 close parentheses comma blank open parentheses x plus 2 close parentheses less than 0 blank a t a u blank x less than negative 2 end cell end table close end cell row cell open vertical bar x plus 2 close vertical bar end cell equals cell open curly brackets table row cell open parentheses x plus 2 close parentheses comma blank open parentheses x plus 2 close parentheses greater or equal than 0 blank a t a u blank x greater or equal than negative 2 end cell row cell negative open parentheses x plus 2 close parentheses comma blank open parentheses x plus 2 close parentheses less than 0 blank a t a u blank x less than negative 2 end cell end table close end cell row cell open vertical bar x minus 1 close vertical bar end cell equals cell open curly brackets table row cell open parentheses x minus 1 close parentheses comma blank open parentheses x minus 1 close parentheses greater or equal than 0 blank a t a u blank x greater or equal than 1 end cell row cell negative open parentheses x minus 1 close parentheses comma blank open parentheses x minus 1 close parentheses less than 0 blank a t a u blank x less than 1 end cell end table close end cell end table end style  

Gabungan daerah untuk interval nilai x di atas menghasilkan 3 daerah sebagai berikut ini.

Daerah I
Untuk interval begin mathsize 14px style x less than negative 2 end style maka diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar negative open parentheses 3 x plus 6 close parentheses minus open square brackets negative open parentheses x plus 2 close parentheses close square brackets close vertical bar end cell less than cell negative open parentheses x minus 1 close parentheses end cell row cell open vertical bar negative 3 x minus 6 plus x plus 2 close vertical bar end cell less than cell negative x plus 1 end cell row cell open vertical bar negative 2 x minus 4 close vertical bar end cell less than cell negative x plus 1 end cell row cell open parentheses negative 2 x minus 4 close parentheses squared end cell less than cell open parentheses negative x plus 1 close parentheses squared end cell row cell 4 x squared plus 16 x plus 16 end cell less than cell x squared minus 2 x plus 1 end cell row cell 4 x squared minus x squared plus 16 x plus 2 x plus 16 minus 1 end cell less than 0 row cell 3 x squared plus 18 x plus 15 end cell less than 0 row cell x squared plus 6 x plus 5 end cell less than 0 row cell open parentheses x plus 5 close parentheses open parentheses x plus 1 close parentheses end cell less than 0 row blank blank blank end table end style   

Irisan daerah himpunan penyelesaian begin mathsize 14px style open parentheses x plus 5 close parentheses open parentheses x plus 1 close parentheses less than 0 end style dan undefined yaitu

Maka, himpunan penyelesaian daerah I adalah begin mathsize 14px style open curly brackets negative 5 less than x less than negative 2 close curly brackets end style 

Daerah II
Untuk interval begin mathsize 14px style negative 2 less or equal than x less than 1 end style maka diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 3 x plus 6 minus open parentheses x plus 2 close parentheses close vertical bar end cell less than cell negative open parentheses x minus 1 close parentheses end cell row cell open vertical bar 3 x plus 6 minus x minus 2 close vertical bar end cell less than cell negative x plus 1 end cell row cell open vertical bar 2 x plus 4 close vertical bar end cell less than cell negative x plus 1 end cell row cell open parentheses 2 x plus 4 close parentheses squared end cell less than cell open parentheses negative x plus 1 close parentheses squared end cell row cell 4 x squared plus 16 x plus 16 end cell less than cell x squared minus 2 x plus 1 end cell row cell 4 x squared minus x squared plus 16 x plus 2 x plus 16 minus 1 end cell less than 0 row cell 3 x squared plus 18 x plus 15 end cell less than 0 row cell x squared plus 6 x plus 5 end cell less than 0 row cell open parentheses x plus 5 close parentheses open parentheses x plus 1 close parentheses end cell less than 0 end table end style   

Irisan daerah himpunan penyelesaian undefined dan undefined yaitu 

Maka, himpunan penyelesaian daerah II adalah begin mathsize 14px style open curly brackets negative 2 less than x less than negative 1 close curly brackets end style 

Daerah III
Untuk interval begin mathsize 14px style x less or equal than 1 end style maka diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 3 x plus 6 minus open parentheses x plus 2 close parentheses close vertical bar end cell less than cell x minus 1 end cell row cell open vertical bar 3 x plus 6 minus x minus 2 close vertical bar end cell less than cell x minus 1 end cell row cell open vertical bar 2 x plus 4 close vertical bar end cell less than cell x minus 1 end cell row cell open parentheses 2 x plus 4 close parentheses squared end cell less than cell open parentheses x minus 1 close parentheses squared end cell row cell 4 x squared plus 16 x plus 16 end cell less than cell x squared minus 2 x plus 1 end cell row cell 4 x squared minus x squared plus 16 x plus 2 x plus 16 minus 1 end cell less than 0 row cell 3 x squared plus 18 x plus 15 end cell less than 0 row cell x squared plus 6 x plus 5 end cell less than 0 row cell open parentheses x plus 5 close parentheses open parentheses x plus 1 close parentheses end cell less than 0 end table end style   

Irisan daerah himpunan penyelesaian undefined dan undefined yaitu

Maka, himpunan penyelesaian daerah III adalah begin mathsize 14px style open curly brackets negative 5 less than x less than negative 1 close curly brackets end style 

Jika kita gabungkan daerah himpunan penyelesaian dari ketiga daerah di atas dalam sebuah garis bilangan diperoleh gambar berikut ini.

Maka, himpunan penyelesaian pertidaksamaan begin mathsize 14px style open vertical bar open vertical bar 3 x plus 6 close vertical bar minus open vertical bar x plus 2 close vertical bar close vertical bar less than open vertical bar x minus 1 close vertical bar end style adalah begin mathsize 14px style open curly brackets x vertical line minus 5 less or equal than x less than negative 1 comma blank x element of R close curly brackets end style 

Jadi, jawaban yang tepat adalah B.

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