Selesaikanlah. b. ∣∣x+52x−1∣∣≤3

Pertanyaan

Selesaikanlah.

b. $open vertical bar fraction numerator 2 x minus 1 over denominator x plus 5 end fraction close vertical bar less or equal than 3$

W. Wati

Master Teacher

Jawaban terverifikasi

Jawaban

himpunan penyelesaiannya adalah H p equals open curly brackets x vertical line x less or equal than negative 16 space atau space x greater or equal than negative 14 over 5 close curly brackets

H p equals open curly brackets x vertical line x less or equal than negative 16 space atau space x greater or equal than negative 14 over 5 close curly brackets

Pembahasan

Pertidaksaman tersebut merupakan pertidaksamaan nilai mutlak, yang fungsinya berbentuk fungsi rasional, syaratnya penyebut tidak boleh sama dengan nol. Untuk mancari penyelesaiannya bisa kita kuadratkan kedua ruas.

$space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open vertical bar fraction numerator 2 x minus 1 over denominator x plus 5 end fraction close vertical bar squared less or equal than 3 squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses fraction numerator 2 x minus 1 over denominator x plus 5 end fraction close parentheses squared less or equal than 3 squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses fraction numerator 2 x minus 1 over denominator x plus 5 end fraction close parentheses squared minus 3 squared less or equal than 0 space space space space space space space space space space space space space space space space space space space space space space open parentheses fraction numerator 2 x minus 1 over denominator x plus 5 end fraction minus 3 close parentheses open parentheses fraction numerator 2 x minus 1 over denominator x plus 5 end fraction plus 3 close parentheses less or equal than 0 open parentheses fraction numerator 2 x minus 1 minus 3 open parentheses x plus 5 close parentheses over denominator x plus 5 end fraction close parentheses open parentheses fraction numerator 2 x minus 1 plus 3 open parentheses x plus 5 close parentheses over denominator x plus 5 end fraction close parentheses less or equal than 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses fraction numerator negative x minus 16 over denominator x plus 5 end fraction close parentheses open parentheses fraction numerator 5 x plus 14 over denominator x plus 5 end fraction close parentheses less or equal than 0$

Pembuat nol pada pembilang: x equals negative 16 space atau space x equals negative 14 over 5 .

Pembuat nol pada penyebut: x equals negative 5 .

Kemudian kita buat garis bilangan, dan kita tentukan tanda disetiap daerahnya dengan cara kita uji titik.

bottom enclose negative negative negative space space space plus plus plus space space space plus plus plus space space space minus negative negative end enclose space space space space space space minus 16 space space space space space space space space minus 5 space space space space space space minus 14 over 5

karena tanda pertidaksamaan kita adalah " less or equal than " , maka pada garis bilangan kita ambil daerah yang bernilai negatif. yaitu,

x less or equal than negative 16 space atau space x greater or equal than negative 14 over 5

Jadi, himpunan penyelesaiannya adalah .