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Jika maka nilai k yang tepat adalah ….

Jika begin mathsize 14px style open parentheses x close parentheses equals fraction numerator 1 over denominator square root of x squared minus sign 2 end root end fraction comma space open parentheses fog close parentheses open parentheses x close parentheses equals fraction numerator 1 over denominator square root of x squared plus 6 x plus 7 end root end fraction space dan space g open parentheses x and k close parentheses equals x plus 5 end style maka nilai k yang tepat adalah ….

  1. 2

  2. -2

  3. -4

  4. -6

  5. 5

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Karena maka: Ambil salah satu persamaan :

begin mathsize 14px style blank end stylebegin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis fog right parenthesis left parenthesis straight x right parenthesis end cell equals cell fraction numerator 1 over denominator square root of straight x squared plus 6 straight x plus 7 end root end fraction end cell row cell straight f left parenthesis straight g left parenthesis straight x right parenthesis right parenthesis end cell equals cell fraction numerator 1 over denominator square root of straight x squared plus 6 straight x plus 7 end root end fraction end cell row cell fraction numerator 1 over denominator square root of straight g left parenthesis straight x right parenthesis squared minus 2 end root end fraction end cell equals cell fraction numerator 1 over denominator square root of straight x squared plus 6 straight x plus 7 end root end fraction end cell row cell straight g left parenthesis straight x right parenthesis squared minus 2 end cell equals cell straight x squared plus 6 straight x plus 7 end cell row cell straight g left parenthesis straight x right parenthesis end cell equals cell square root of straight x squared plus 6 straight x plus 9 end root end cell end table end style   


Karena begin mathsize 14px style g open parentheses x and k close parentheses equals x plus 5 end style maka:

undefinedbegin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight g left parenthesis straight x plus straight k right parenthesis end cell equals cell square root of left parenthesis straight x plus straight k right parenthesis squared plus 6 left parenthesis straight x plus straight k right parenthesis plus 9 end root end cell row cell straight x plus 5 end cell equals cell square root of straight x squared plus 2 kx plus straight k squared plus 6 straight x plus 6 straight k plus 9 end root end cell row cell left parenthesis straight x plus 5 right parenthesis squared end cell equals cell straight x to the power of 2 end exponent plus left parenthesis 2 straight k plus 6 right parenthesis straight x plus left parenthesis straight k squared plus 6 straight k plus 9 right parenthesis end cell row cell straight x squared plus 10 straight x plus 25 end cell equals cell straight x squared plus left parenthesis 2 straight k plus 6 right parenthesis straight x plus left parenthesis straight k squared plus 6 straight k plus 9 right parenthesis end cell end table end style    

Ambil salah satu persamaan :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 10 straight x end cell equals cell left parenthesis 2 straight k plus 6 right parenthesis straight x end cell row 10 equals cell 2 straight k plus 6 end cell row 4 equals cell 2 straight k end cell row 2 equals straight k end table end style   

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Diketahui h ( x ) = 2 x − 4 , f ( x ) = x 2 − 4 , dan ( h ∘ g ∘ f ) ( x ) = x 2 . Jika , maka nilai k yang memenuhi adalah ....

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