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Himpunan penyelesaian dari pertidaksamaan 5 x − 2 8 x 2 − 3 x + 10 ​ ≤ 2 x − 1 adalah ...

Himpunan penyelesaian dari pertidaksamaan  adalah ...

  1. left curly bracket x less or equal than negative 1 space a t a u space 2 over 3 less than x less or equal than 4 right curly bracket

  2. left curly bracket negative 1 less or equal than x less than 2 over 5 space space a t a u space x greater or equal than 4 right curly bracket

  3. open curly brackets x less or equal than negative 1 space a t a u italic space italic x greater or equal than 4 close curly brackets

  4. open curly brackets x less than 2 over 5 a t a u italic x greater or equal than 4 close curly brackets

  5. open curly brackets x less or equal than negative 1 a t a u italic x greater or equal than 4 close curly brackets

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N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

himpunan penyelesaiannya adalah

himpunan penyelesaiannya adalah open curly brackets x vertical line minus 1 less or equal than x less than 2 over 5 a t a u italic x greater or equal than 4 close curly brackets

Pembahasan

Jadi, himpunan penyelesaiannya adalah

fraction numerator 8 x squared minus 3 x plus 10 over denominator 5 x minus 2 end fraction less or equal than 2 x minus 1 fraction numerator 8 x squared minus 3 x plus 10 over denominator 5 x minus 2 end fraction minus open parentheses 2 x minus 1 close parentheses less or equal than 0 fraction numerator 8 x squared minus 3 x plus 10 minus open parentheses 2 x minus 1 close parentheses open parentheses 5 x minus 2 close parentheses over denominator 5 x minus 2 end fraction less or equal than 0 fraction numerator 8 x squared minus 3 x plus 10 minus open parentheses 10 x squared minus 9 x plus 2 close parentheses over denominator 5 x minus 2 end fraction less or equal than 0 fraction numerator 8 x squared minus 3 x plus 10 minus 10 x squared plus 9 x minus 2 over denominator 5 x minus 2 end fraction less or equal than 0 fraction numerator negative 2 x squared plus 6 x plus 8 over denominator 5 x minus 2 end fraction less or equal than 0 fraction numerator negative 2 open parentheses x minus 4 close parentheses open parentheses x plus 1 close parentheses over denominator 5 x minus 2 end fraction less or equal than 0  P e m b u a t space n o l space n y a space a d a l a h x equals 4 comma space x equals negative 1 comma space d a n space x equals 2 over 5

 

Jadi, himpunan penyelesaiannya adalah open curly brackets x vertical line minus 1 less or equal than x less than 2 over 5 a t a u italic x greater or equal than 4 close curly brackets

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