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Himpunan penyelesaian pertidaksamaan  adalah ...

Himpunan penyelesaian pertidaksamaan begin mathsize 14px style fraction numerator x squared plus 2 x minus 3 over denominator 6 minus 3 x minus 3 x squared end fraction less or equal than 2 end style adalah ...

  1. begin mathsize 14px style open curly brackets x vertical line x less or equal than negative 15 over 7 space atau space x greater than negative 2 comma space x not equal to 1 close curly brackets end style   

  2. begin mathsize 14px style open curly brackets x vertical line x less or equal than negative 15 over 7 space atau space minus 2 less than space x less than 1 close curly brackets end style 

  3. begin mathsize 14px style open curly brackets x vertical line minus 15 over 7 less or equal than x less or equal than negative 2 space atau space x greater than 1 close curly brackets end style 

  4. begin mathsize 14px style open curly brackets x vertical line x less or equal than negative 15 over 7 space atau space x greater than negative 2 close curly brackets end style 

  5. begin mathsize 14px style open curly brackets x vertical line minus 15 over 7 less or equal than x less or equal than negative 2 close curly brackets end style 

Jawaban:

Untuk menyelesaikan pertidaksamaan tersebut, jadikan ruas kanan sama dengan 0 seperti berikut:

begin mathsize 14px style table row blank cell fraction numerator x squared plus 2 x minus 3 over denominator 6 minus 3 x minus 3 x squared end fraction less or equal than 2 end cell row left right double arrow cell fraction numerator x squared plus 2 x minus 3 over denominator 6 minus 3 x minus 3 x squared end fraction minus 2 less or equal than 0 end cell row left right double arrow cell fraction numerator x squared plus 2 x minus 3 over denominator 3 x squared plus 3 x minus 6 end fraction minus fraction numerator 2 open parentheses 6 minus 3 x minus 3 x squared close parentheses over denominator 3 x squared plus 3 x minus 6 end fraction greater or equal than 0 end cell row left right double arrow cell fraction numerator 7 x squared plus 8 x minus 15 over denominator 3 x squared plus 3 x minus 6 end fraction greater or equal than 0 end cell row left right double arrow cell fraction numerator open parentheses 7 x plus 15 close parentheses open parentheses x minus 1 close parentheses over denominator 3 open parentheses x plus 2 close parentheses open parentheses x minus 1 close parentheses end fraction greater or equal than 0 end cell row left right double arrow cell fraction numerator 7 x plus 15 over denominator 3 open parentheses x plus 2 close parentheses end fraction greater or equal than 0 end cell end table end style 

Kemudian tentukan pembuat nol:

begin mathsize 14px style table row cell 7 x plus 15 equals 0 end cell rightwards arrow cell x equals negative 15 over 7 end cell row cell x plus 2 equals 0 end cell rightwards arrow cell x equals negative 2 end cell end table end style 

Oleh karena, pertidaksamaan tersebut adalah pertidaksamaan rasional maka syarat:

begin mathsize 14px style table row cell x plus 2 not equal to 0 end cell rightwards arrow cell x not equal to negative 2 end cell row cell x minus 1 not equal to 0 end cell rightwards arrow cell x not equal to 1 end cell end table end style

Selanjutnya buatlah garis bilangan dan ujilah titik paling kanan, sehingga diperoleh garis bilangan berikut:

Dengan demikian, himpunan penyelesaian pertidaksamaan tersebut adalah

begin mathsize 14px style open curly brackets x vertical line x less or equal than negative 15 over 7 space atau space x greater than negative 2 comma space x not equal to 1 close curly brackets end style

Oleh karena itu, jawaban yang benar adalah A.

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