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Himpunan penyelesaian dari pertidaksamaan ∣ x + 3 ∣ ≥ 15 − 2 ∣ x + 3 ∣ adalah ...

Himpunan penyelesaian dari pertidaksamaan  adalah ...

  1. begin mathsize 14px style open curly brackets right enclose x space minus 8 less or equal than x less or equal than negative 2 comma space x element of straight real numbers space close curly brackets end style 

  2. begin mathsize 14px style left curly bracket right enclose x space minus 8 less or equal than x less or equal than 2 comma space x element of straight real numbers space right curly bracket end style 

  3. begin mathsize 14px style open curly brackets right enclose x space x less or equal than negative 8 space atau space x greater or equal than negative 2 comma space x element of straight real numbers space close curly brackets end style 

  4. begin mathsize 14px style left curly bracket right enclose x space x less or equal than negative 8 space atau space x greater or equal than 2 comma space x element of straight real numbers space right curly bracket end style 

  5. begin mathsize 14px style left curly bracket right enclose x space x less or equal than negative 2 space atau space x greater or equal than 8 comma space x element of straight real numbers space right curly bracket end style 

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M. Mariyam

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

Jawaban terverifikasi

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Pembahasan

Himpunan penyelesaian dari pertidaksamaan adalah Kita harus mengingat sifat pertidaksamaan mutlak berikut: Dengan demikian, Jadi, jawaban yang tepat adalah D.

Himpunan penyelesaian dari pertidaksamaan begin mathsize 14px style open vertical bar x plus 3 close vertical bar greater or equal than 15 minus 2 open vertical bar x plus 3 close vertical bar end style adalah

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x plus 3 close vertical bar end cell greater or equal than cell 15 minus 2 open vertical bar x plus 3 close vertical bar end cell row cell open vertical bar x plus 3 close vertical bar plus 2 open vertical bar x plus 3 close vertical bar end cell greater or equal than 15 row cell 3 open vertical bar x plus 3 close vertical bar end cell greater or equal than 15 row cell open vertical bar x plus 3 close vertical bar end cell greater or equal than cell 15 over 3 end cell row cell open vertical bar x plus 3 close vertical bar end cell greater or equal than 5 end table end style 

Kita harus mengingat sifat pertidaksamaan mutlak berikut:

begin mathsize 14px style open vertical bar f open parentheses x close parentheses close vertical bar greater or equal than a left right arrow f open parentheses x close parentheses less or equal than negative a space atau space f open parentheses x close parentheses greater or equal than a end style

Dengan demikian,

begin mathsize 14px style open vertical bar x plus 3 close vertical bar greater or equal than 5 left right arrow open curly brackets right enclose x x plus 3 less or equal than negative 5 space atau space x plus 3 greater or equal than 5 space comma space x element of straight real numbers close curly brackets left right arrow open curly brackets right enclose x x less or equal than negative 5 minus 3 space atau space x greater or equal than 5 minus 3 comma space x element of straight real numbers close curly brackets left right arrow open curly brackets right enclose x x less or equal than negative 8 space space atau space x greater or equal than 2 comma space x element of straight real numbers close curly brackets end style 

Jadi, jawaban yang tepat adalah D.

 

 

 

 

 

Latihan Bab

Konsep Kilat

Nilai Mutlak

Fungsi Nilai Mutlak

Persamaan Nilai Mutlak Linear I

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