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Tentukan himpunan penyelesaian dari !

Tentukan himpunan penyelesaian dari open vertical bar x plus 10 close vertical bar minus 4 less or equal than 0!

Jawaban:

Pertidaksamaan nilai mutlak open vertical bar x plus 10 close vertical bar minus 4 less or equal than 0 diubah menjadi:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x plus 10 close vertical bar minus 4 end cell less or equal than 0 row cell open vertical bar x plus 10 close vertical bar end cell less or equal than 4 end table

Kemudian ingatlah sifat pertidaksamaan nilai mutlak, yaitu:

open vertical bar f open parentheses x close parentheses close vertical bar less or equal than straight a left right double arrow negative straight a less or equal than f open parentheses x close parentheses less or equal than straight a

Sehingga diperoleh:

table attributes columnalign right center center center left end attributes row cell negative 4 end cell less or equal than cell x plus 10 end cell less or equal than 4 row cell negative 4 minus 10 end cell less or equal than x less or equal than cell 4 minus 10 end cell row cell negative 14 end cell less or equal than x less or equal than cell negative 6 end cell end table

Maka, himpunan penyelesaian dari open vertical bar x plus 10 close vertical bar minus 4 less or equal than 0 adalah HP equals open curly brackets right enclose x space minus 14 less or equal than x less or equal than negative 6 comma space x element of straight real numbers close curly brackets.

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