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Jumlah nilai  yang memenuhi sistem persamaan   ad...

Jumlah nilai begin mathsize 14px style x end style yang memenuhi sistem persamaan 

begin mathsize 14px style open curly brackets table attributes columnalign left end attributes row cell x squared plus x y minus y squared equals negative 4 end cell row cell x plus 2 y equals 2 end cell end table close end style adalah ...

  1. begin mathsize 14px style 8 end style 

  2. begin mathsize 14px style 4 end style 

  3. begin mathsize 14px style 1 half end style 

  4. begin mathsize 14px style negative 1 half end style 

  5. begin mathsize 14px style negative 8 end style 

Jawaban:

begin mathsize 14px style open curly brackets table attributes columnalign left end attributes row cell x squared plus x y minus y squared equals negative 4 end cell row cell x plus 2 y equals 2 end cell end table close x plus 2 y equals 2 space rightwards arrow space y equals fraction numerator 2 minus x over denominator 2 end fraction substitusi space y space ke space persamaan space 1. space space space space space space space x squared plus x open parentheses fraction numerator 2 minus x over denominator 2 end fraction close parentheses minus open parentheses fraction numerator 2 minus x over denominator 2 end fraction close parentheses squared equals negative 4 space x squared plus fraction numerator 2 x minus x squared over denominator 2 end fraction minus open parentheses fraction numerator 4 minus 4 x plus x squared over denominator 4 end fraction close parentheses equals negative 4 4 x squared plus 4 x minus 2 x squared minus open parentheses 4 minus 4 x plus x squared close parentheses equals negative 4 space space space space space space space space space space space space 2 x squared plus 4 x minus 4 plus 4 x minus x squared equals negative 4 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x squared plus 8 x equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x open parentheses x plus 8 close parentheses equals 0 space space space space space space space space space space space space space space space space x subscript 1 equals 0 space atau space x subscript 2 equals negative 8 x subscript 1 plus x subscript 2 equals negative 8 end style 

Jadi, jawaban yang tepat adalah E.  

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