Roboguru

Pertanyaan

table attributes columnalign left end attributes row cell straight P subscript 1 space space space space space space end subscript colon space Jika italic space x to the power of italic 2 less or equal than 4 comma space maka space minus 2 less or equal than x less or equal than 2 end cell row cell bottom enclose straight P subscript 2 space space space space space end subscript space colon space x less than negative 2 space atau italic space x greater than 2 space space space space space space space space space space space space space space space space space space space space end enclose end cell end table Konklusi colon space...

Pembahasan Soal:

Misal:

  p equals x squared less or equal than 4

q equals negative 2 less or equal than x less or equal than 2 

Kemudian, negasi dari q yaitu tilde q equals tilde open parentheses negative 2 less or equal than x less or equal than 2 close parentheses equals x less than negative 2 space atau space x greater than 2.

Diketahui:

table attributes columnalign left end attributes row cell straight P subscript 1 space space space space space space end subscript colon space Jika italic space x to the power of italic 2 less or equal than 4 comma space maka space minus 2 less or equal than x less or equal than 2 end cell row cell bottom enclose straight P subscript 2 space space space space space end subscript space colon space x less than negative 2 space atau italic space x greater than 2 space space space space space space space space space space space space space space space space space space space space end enclose end cell end table Konklusi colon space...

Kedua premis di atas dapat dituliskan sebagai berikut.

straight P subscript 1 space space space colon space p rightwards double arrow q straight P subscript 2 space space space colon tilde q

Berdasarkan prinsip penarikan kesimpulan modus tollens, kesimpulan dari kedua premis di atas yaitu tilde p. Negasi dari p yaitu tilde p equals tilde open parentheses x squared less or equal than 4 close parentheses equals x squared greater than 4

Dengan demikian, kesimpulan dari premis di atas sebagai berikut.

 table attributes columnalign left end attributes row cell straight P subscript 1 space space space space space space end subscript colon space Jika italic space x to the power of italic 2 less or equal than 4 comma space maka space minus 2 less or equal than x less or equal than 2 end cell row cell bottom enclose straight P subscript 2 space space space space space end subscript space colon space x less than negative 2 space atau italic space x greater than 2 space space space space space space space space space space space space space space space space space space space space end enclose end cell end table Konklusi colon space x squared greater than 4

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 06 Mei 2021

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