Roboguru

Tentukan himpunan penyelesaian dari persamaan dan pertidaksaman nilai mutlak berikut. c.   td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}

Pertanyaan

Tentukan himpunan penyelesaian dari persamaan dan pertidaksaman nilai mutlak berikut.

c. begin mathsize 14px style open vertical bar x over 3 minus 2 close vertical bar less or equal than 10 end style  td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}

  1. ... undefined 

  2. ... undefined  

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x over 3 minus 2 close vertical bar space end cell rightwards arrow cell open parentheses x over 3 minus 2 close parentheses space j i k a space x greater or equal than 6 end cell row cell space space space space space space space space space space space end cell rightwards arrow cell open parentheses 2 minus x over 3 close parentheses space j i k a space x less than 6 end cell row blank blank blank row cell Kondisi space 1 space colon space x end cell greater or equal than 6 row cell x over 3 minus 2 end cell less or equal than 10 row cell x over 3 minus 12 end cell less or equal than 0 row cell fraction numerator x minus 36 over denominator 3 end fraction end cell less or equal than 0 row x less or equal than 36 row blank blank cell Diiriskan space dengan space syarat space kondisi space end cell row x greater or equal than cell 6 space end cell row blank blank cell Sehingga space irisannya space menjadi space end cell row 6 less or equal than cell x less or equal than 36 end cell row blank blank blank row cell Kondisi space 2 space colon space x end cell less than 6 row cell 2 minus x over 3 end cell less or equal than 10 row cell fraction numerator negative x over denominator 3 end fraction end cell less or equal than 8 row cell negative x end cell less or equal than 24 row x greater or equal than cell negative 24 end cell row blank blank cell Diiriskan space dengan space syarat space kondisi space 2 colon end cell row cell negative 24 end cell less or equal than cell x less than 6 end cell row blank blank blank row blank blank blank row blank blank space end table end style

 

Jadi daerah gabungan penyelesaian adalah :

begin mathsize 14px style negative 24 less or equal than x less or equal than 36 end style  

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 08 Februari 2021

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