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Carilah penyelesaian dari setiap sistem pertidaksamaan di bawah ini. a. {x+22​−x−31​−x+51​≥0x−11​≤2x+11​​

Pertanyaan

Carilah penyelesaian dari setiap sistem pertidaksamaan di bawah ini.

a. open curly brackets table attributes columnalign left end attributes row cell fraction numerator 2 over denominator x plus 2 end fraction minus fraction numerator 1 over denominator x minus 3 end fraction minus fraction numerator 1 over denominator x plus 5 end fraction greater or equal than 0 end cell row cell fraction numerator 1 over denominator x minus 1 end fraction less or equal than fraction numerator 1 over denominator 2 x plus 1 end fraction end cell end table close 

S. Yoga

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

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didapat irisan penyelesaian pertidaksamaan tersebut adalah negative 17 less or equal than x less than negative 5 space atau space minus 1 half less than straight x less than 1.

Pembahasan

Cara menyelesaikan pertidaksamaan rasional linear kuadrat :

  1.  Jadikan ruas kanan = 0.
  2. Ubah tanda koefisien variabel x squared pada bentuk kuadrat dan koefisien x pada bentuk linear menjadi bertanda sama.
  3. Carilah nilai nol pembilang maupun penyebut.
  4. Pembilang atau penyebut yang berbentuk kuadrat difaktorkan terlebih dahulu.
  5. Buat garis bilangan untuk menentukan interval atau batas penyelesaian.

Pertidaksamaan fraction numerator 2 over denominator x plus 2 end fraction minus fraction numerator 1 over denominator x minus 3 end fraction minus fraction numerator 1 over denominator x plus 5 end fraction greater or equal than 0:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 over denominator x plus 2 end fraction minus fraction numerator 1 over denominator x minus 3 end fraction minus fraction numerator 1 over denominator x plus 5 end fraction end cell greater or equal than cell thin space 0 end cell row cell fraction numerator negative 2 x minus 34 over denominator open parentheses x plus 2 close parentheses open parentheses x minus 3 close parentheses open parentheses x plus 5 close parentheses end fraction end cell greater or equal than 0 row cell fraction numerator negative 2 open parentheses x plus 17 close parentheses over denominator open parentheses x plus 2 close parentheses open parentheses x minus 3 close parentheses open parentheses x plus 5 close parentheses end fraction end cell greater or equal than 0 row cell fraction numerator 2 open parentheses x plus 17 close parentheses over denominator open parentheses x plus 2 close parentheses open parentheses x minus 3 close parentheses open parentheses x plus 5 close parentheses end fraction end cell less or equal than 0 row cell fraction numerator x plus 17 over denominator open parentheses x plus 2 close parentheses open parentheses x minus 3 close parentheses open parentheses x plus 5 close parentheses end fraction end cell less or equal than 0 end table    

Titik nol :

    table attributes columnalign right center left columnspacing 0px end attributes row cell Pembilang space colon space x plus 17 end cell equals 0 row blank rightwards arrow cell x equals negative 17 end cell row cell Penyebut space colon space open parentheses x plus 2 close parentheses open parentheses x minus 3 close parentheses open parentheses x plus 5 close parentheses end cell not equal to 0 row blank rightwards arrow cell x not equal to negative 2 comma space end cell row x not equal to cell 3 comma space end cell row x not equal to cell negative 5 end cell end table  

Garis bilangan:



Sehingga penyelesaian pertidaksamaan tersebut adalah negative 17 less or equal than x less than negative 5 space atau space minus 2 less than x less than 3.

Pertidaksamaan fraction numerator 1 over denominator x minus 1 end fraction less or equal than fraction numerator 1 over denominator 2 x plus 1 end fraction:

      table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 over denominator x minus 1 end fraction end cell less or equal than cell fraction numerator 1 over denominator 2 x plus 1 end fraction end cell row cell fraction numerator 1 over denominator x minus 1 end fraction minus fraction numerator 1 over denominator 2 x plus 1 end fraction end cell less or equal than 0 row cell fraction numerator x plus 2 over denominator open parentheses x minus 1 close parentheses open parentheses 2 x plus 1 close parentheses end fraction end cell less or equal than 0 end table  

Titik nol :

    Pembilang space colon space x plus 2 equals 0 rightwards arrow x equals negative 2 Penyebut space colon space open parentheses x minus 1 close parentheses open parentheses 2 x plus 1 close parentheses not equal to 0 rightwards arrow x not equal to 1 comma space x not equal to negative 1 half   

Garis bilangan:



Sehingga penyelesaian pertidaksamaan tersebut adalah x less or equal than negative 2 space atau thin space minus 1 half less than x less than 1.

Dari penyelesaian:

  1. negative 17 less or equal than x less than negative 5 space atau space minus 2 less than x less than 3
  2. x less or equal than negative 2 space atau thin space minus 1 half less than x less than 1

Jadi, didapat irisan penyelesaian pertidaksamaan tersebut adalah negative 17 less or equal than x less than negative 5 space atau space minus 1 half less than straight x less than 1.

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