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Himpunan penyelesaian pertidaksamaan  adalah ...

Himpunan penyelesaian pertidaksamaan begin mathsize 14px style fraction numerator 4 x plus 7 over denominator x plus 3 end fraction greater than negative 1 end style adalah ...

  1. begin mathsize 14px style open curly brackets x vertical line minus 3 less than x less than negative 2 comma space x element of straight R close curly brackets end style  

  2. begin mathsize 14px style open curly brackets x vertical line minus 2 less than x less than negative 3 comma space x element of straight R close curly brackets end style 

  3. begin mathsize 14px style open curly brackets x vertical line x less than negative 3 space atau space x greater than negative 2 comma space x element of straight R close curly brackets end style 

  4. begin mathsize 14px style open curly brackets x vertical line x less than negative 3 space atau space x greater than 2 comma space x element of straight R close curly brackets end style 

  5. begin mathsize 14px style open curly brackets x vertical line x less than negative 2 space atau space x greater than 3 comma space x element of straight R 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 4 x plus 7 over denominator x plus 3 end fraction greater than negative 1 end cell row left right double arrow cell fraction numerator 4 x plus 7 over denominator x plus 3 end fraction plus 1 greater than 0 end cell row left right double arrow cell fraction numerator 4 x plus 7 over denominator x plus 3 end fraction plus fraction numerator x plus 3 over denominator x plus 3 end fraction greater than 0 end cell row left right double arrow cell fraction numerator 5 x plus 10 over denominator x plus 3 end fraction greater than 0 end cell end table end style 

Tentukan pembuat nol:

begin mathsize 14px style table row cell 5 x plus 10 equals 0 end cell rightwards arrow cell x equals negative 2 end cell row cell x plus 3 equals 0 end cell rightwards arrow cell x equals negative 3 end cell end table end style

karena termasuk pertidaksamaan rasional maka syarat:

begin mathsize 14px style table row cell x plus 3 not equal to 0 end cell rightwards arrow cell x not equal to negative 3 end cell end table end style

Oleh karena tanda pertidaksamaan lebih dari begin mathsize 14px style open parentheses greater than close parentheses end style maka untuk interval begin mathsize 14px style x greater than negative 2 end style ambil begin mathsize 14px style x equals 10 end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 4 open parentheses 10 close parentheses plus 7 over denominator 10 plus 3 end fraction end cell equals cell 47 over 13 end cell row blank equals cell 3 comma 62 greater than negative 1 space open parentheses plus close parentheses end cell end table end style 

Sehingga diperoleh garis bilangan sebagai berikut:

Dengan demikian, himpunan penyelesaian dari pertidaksamaan tersebut adalah

begin mathsize 14px style open curly brackets x vertical line x less than negative 3 space atau space x greater than negative 2 comma space x element of straight R close curly brackets end style.

Oleh karena itu, jawaban yang benar adalah C.

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