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Himpunan penyelesaian dari (21​)8+2x−x2>(21​)x+2 adalah .....

Pertanyaan

Himpunan penyelesaian dari open parentheses 1 half close parentheses to the power of 8 plus 2 x minus x squared end exponent greater than open parentheses 1 half close parentheses to the power of x plus 2 end exponent adalah .....

  1. open curly brackets right enclose x x less than negative 2 space atau space x greater than 5 close curly brackets     

  2. open curly brackets right enclose x x less than negative 2 space atau space x greater than 3 close curly brackets     

  3. open curly brackets right enclose x x less than negative 3 space atau space x greater than 2 close curly brackets  

  4. open curly brackets right enclose x x less than negative 2 space atau space x less than 3 close curly brackets  

  5. open curly brackets right enclose x x less than negative 3 space atau space x less than negative 5 close curly brackets  

A. Septianingsih

Master Teacher

Mahasiswa/Alumni Universitas Gadjah Mada

Jawaban terverifikasi

Pembahasan

Ingat rumus pertidaksamaan eksponensial.

table attributes columnalign right center left columnspacing 0px end attributes row cell a to the power of f left parenthesis x right parenthesis end exponent end cell greater than cell a to the power of g left parenthesis x right parenthesis end exponent end cell row cell jika space 0 end cell less than cell a less than 1 space maka space colon end cell row cell f left parenthesis x right parenthesis end cell less than cell g left parenthesis x right parenthesis end cell end table 

Berdasarkan rumus di atas maka berlaku.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 1 half close parentheses to the power of 8 plus 2 x minus x squared end exponent end cell greater than cell open parentheses 1 half close parentheses to the power of x plus 2 end exponent space space colon end cell row cell 8 plus 2 straight x minus straight x squared end cell less than cell straight x plus 2 end cell row cell negative straight x squared plus straight x plus 6 end cell less than 0 row cell straight x squared minus straight x minus 6 end cell greater than 0 row cell open parentheses straight x minus 3 close parentheses open parentheses straight x plus 2 close parentheses end cell greater than 0 row cell titik space batas space straight x end cell equals cell 3 space dan space straight x equals negative 2 end cell end table  

selanjutnya akan dicari daerah penyelesaian dengan menggunakan garis bilangan :

Berdasarkan garis bilangan persamaan kuadrat maka diperoleh penyelesaian open curly brackets right enclose x x less than negative 2 space atau space x greater than 3 close curly brackets.

Oleh karena itu jawaban yang tepat adalah B.

 

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