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Pertanyaan

Himpunan penyelesaian dari pertidaksamaan eksponen fourth root of open parentheses 1 third close parentheses to the power of 2 x minus 1 end exponent end root less or equal than open parentheses 1 over 27 close parentheses to the power of x plus 1 end exponent adalah ....

  1. open curly brackets x vertical line x greater or equal than 7 1 half close curly brackets

  2. open curly brackets x vertical line x greater or equal than 3 over 10 close curly brackets

  3. open curly brackets x vertical line x greater or equal than negative 13 over 10 close curly brackets

  4. open curly brackets x vertical line 2 less than x less than 4 close curly brackets

  5. open curly brackets x vertical line x less or equal than negative 2 space a t a u space x greater or equal than 5 close curly brackets

F. Pratama

Master Teacher

Mahasiswa/Alumni Universitas Putra Indonesia YPTK Padang

Jawaban terverifikasi

Pembahasan

fourth root of open parentheses 1 third close parentheses to the power of 2 x minus 1 end exponent end root less or equal than open parentheses 1 over 27 close parentheses to the power of x plus 1 end exponent  open parentheses 1 third close parentheses to the power of 2 x minus 1 end exponent less or equal than open parentheses 1 third close parentheses to the power of 3 x plus 3 end exponent  1 half x minus 1 fourth greater or equal than 3 x plus 3  2 1 half x greater or equal than negative 3 1 fourth  5 over 2 x greater or equal than negative 13 over 4  x greater or equal than negative 13 over 10

HP=open curly brackets x vertical line x greater or equal than negative 13 over 10 close curly brackets

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