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Himpunan penyelesaian dari pertidaksamaan fraction numerator x squared minus 3 open vertical bar x close vertical bar plus 2 over denominator x plus 1 end fraction greater or equal than 0 adalah ... 

  1. open curly brackets x element of R vertical line minus 1 less than x less or equal than 1 space atau space x greater or equal than 2 close curly brackets   

  2. open curly brackets x element of R vertical line minus 2 less than x less than negative 1 space atau space x greater or equal than 2 close curly brackets  

  3. open curly brackets x element of R vertical line minus 2 less or equal than x less than negative 1 space atau space minus 1 less than x less or equal than 1 close curly brackets 

  4. open curly brackets x element of R vertical line minus 2 less or equal than x less than negative 1 space atau space minus 1 less than x less or equal than 1 space atau space x greater or equal than 2 close curly brackets 

  5. open curly brackets x element of R vertical line x less than negative 1 space atau space minus 1 less than x less or equal than 1 space atau space x greater or equal than 2 close curly brackets 

L. Rante

Master Teacher

Mahasiswa/Alumni Universitas Negeri Makassar

Jawaban terverifikasi

Pembahasan

Ingat! 

  • Pada pertidaksamaan pecahan penyebut not equal to 0 
  • open vertical bar x close vertical bar equals c rightwards double arrow x equals plus-or-minus c 

Perhatikan perhitungan di bawah ini! 

fraction numerator x squared minus 3 open vertical bar x close vertical bar plus 2 over denominator x plus 1 end fraction greater or equal than 0

Pembuat nol pembilang 

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 3 open vertical bar x close vertical bar plus 2 end cell equals 0 row cell open vertical bar x close vertical bar squared minus 3 open vertical bar x close vertical bar plus 2 end cell equals 0 row cell open parentheses open vertical bar x close vertical bar minus 1 close parentheses open parentheses open vertical bar x close vertical bar minus 2 close parentheses end cell equals 0 row cell open vertical bar x close vertical bar minus 1 end cell equals cell 0 space space atau space open vertical bar x close vertical bar minus 2 equals 0 end cell row cell open vertical bar x close vertical bar end cell equals cell 1 space space space space space space space space space space space space space space space open vertical bar x close vertical bar equals 2 end cell row x equals cell plus-or-minus 1 space space space space space space space space space space space space space space x equals plus-or-minus 2 end cell end table 

Penyebut tidak boleh bernilai 0, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 1 end cell not equal to 0 row x not equal to cell negative 1 end cell end table 

Garis bilangan 


 


Jadi, himpunan penyelesaian dari pertidaksamaan tersebut adalah open curly brackets x element of R vertical line minus 2 less or equal than x less or equal than negative 1 space atau space minus 1 less than x less or equal than 1 space atau space x greater or equal than 2 close curly brackets

Oleh karena itu, jawaban yang benar adalah D.

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