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Tentukan penyelesaian pertidaksamaan pecahan berik...

Tentukan penyelesaian pertidaksamaan pecahan berikut!

begin mathsize 14px style straight a. space fraction numerator x plus 1 over denominator x minus 3 end fraction less than 4 end style

Jawaban:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x plus 1 over denominator x minus 3 end fraction end cell less than 4 row cell fraction numerator x plus 1 over denominator x minus 3 end fraction minus 4 end cell less than 0 row cell fraction numerator x plus 1 minus 4 x plus 12 over denominator x minus 3 end fraction end cell less than 0 row cell fraction numerator negative 3 x plus 13 over denominator x minus 3 end fraction end cell less than 0 row blank blank blank row cell negative 3 x plus 13 end cell equals 0 row cell negative 3 x end cell equals cell negative 13 end cell row x equals cell 13 over 3 end cell row blank blank blank row cell x minus 3 end cell equals 0 row x equals 3 row blank blank blank row cell x minus 3 end cell not equal to 0 row x not equal to 3 end table end style

Jadi, begin mathsize 14px style x less than 3 space atau space x greater than 13 over 3 end style 

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