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Carilah himpunan penyelesaian dari setiap pertidaksamaan berikut. b.

Pertanyaan

Carilah himpunan penyelesaian dari setiap pertidaksamaan berikut.

b. fraction numerator negative 2 open parentheses 5 minus 6 x close parentheses over denominator 2 end fraction plus 3 less or equal than negative 7 over 2 plus 14 x 

Pembahasan Soal:

b.  Dengan menggunakan sifat-sifat dasar pertidaksamaan, penyelesaian dari pertidaksamaan di atas sebagai berikut.
 

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator negative 2 open parentheses 5 minus 6 x close parentheses over denominator 2 end fraction plus 3 end cell less or equal than cell negative 7 over 2 plus 14 x end cell row cell fraction numerator negative 10 plus 12 x over denominator 2 end fraction plus 3 end cell less or equal than cell negative 7 over 2 plus 14 x end cell row cell fraction numerator negative 10 plus 12 x over denominator 2 end fraction plus 3 minus 3 end cell less or equal than cell negative 7 over 2 plus 14 x minus 3 end cell row cell fraction numerator negative 10 plus 12 x over denominator 2 end fraction end cell less or equal than cell negative 7 over 2 minus 6 over 2 plus 14 x end cell row cell fraction numerator negative 10 plus 12 x over denominator 2 end fraction end cell less or equal than cell negative 13 over 2 plus 14 x end cell row cell fraction numerator negative 10 plus 12 x over denominator 2 end fraction minus 14 x end cell less or equal than cell negative 13 over 2 plus 14 x minus 14 x end cell row cell fraction numerator open parentheses negative 10 plus 12 x close parentheses over denominator 2 end fraction minus fraction numerator 14 x times 2 over denominator 2 end fraction end cell less or equal than cell negative 13 over 2 end cell row cell fraction numerator negative 10 plus 12 x minus 28 x over denominator 2 end fraction end cell less or equal than cell negative 13 over 2 end cell row cell fraction numerator negative 10 minus 16 x over denominator 2 end fraction end cell less or equal than cell negative 13 over 2 end cell row cell fraction numerator negative 10 minus 16 x over denominator 2 end fraction times 2 end cell less or equal than cell negative 13 over 2 times 2 end cell row cell negative 10 minus 16 x end cell less or equal than cell negative 13 end cell row cell negative 10 minus 16 x plus 10 end cell less or equal than cell negative 13 plus 10 end cell row cell negative 16 x end cell less or equal than cell negative 3 end cell row cell negative 16 x times open parentheses negative 1 over 16 close parentheses end cell greater or equal than cell negative 3 times open parentheses negative 1 over 16 close parentheses end cell row x greater or equal than cell 3 over 16 end cell end table 

 

Dengan demikian, himpunan penyelesaian dari fraction numerator negative 2 open parentheses 5 minus 6 x close parentheses over denominator 2 end fraction plus 3 less or equal than negative 7 over 2 plus 14 x adalah open curly brackets x vertical line space x greater or equal than 3 over 16 close curly brackets.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 10 Juni 2021

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