Roboguru

Tentukan interval nilai  yang memenuhi pertidaksamaan berikut ini ; b.

Pertanyaan

Tentukan interval nilai x yang memenuhi pertidaksamaan berikut ini ;

b. table attributes columnalign right center left columnspacing 0px end attributes row cell size 12px vertical line size 12px 2 size 12px x size 12px minus size 12px 1 size 12px vertical line end cell size 12px less or equal than cell size 12px vertical line size 12px x size 12px minus size 12px 5 size 12px vertical line end cell end table

Pembahasan Soal:

Ingat!

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a close vertical bar end cell less or equal than cell open vertical bar b close vertical bar end cell row cell open vertical bar a close vertical bar squared end cell less or equal than cell open vertical bar b close vertical bar squared end cell row cell a squared end cell less or equal than cell b squared end cell row cell a squared minus b squared end cell less or equal than 0 row cell left parenthesis a minus b right parenthesis left parenthesis a plus b right parenthesis end cell less or equal than 0 end table  

Maka:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell vertical line 2 x minus 1 vertical line end cell less or equal than cell vertical line x minus 5 vertical line end cell row cell left parenthesis 2 x minus 1 right parenthesis squared end cell less or equal than cell left parenthesis x minus 5 right parenthesis squared space end cell row cell left parenthesis 2 x minus 1 right parenthesis squared minus left parenthesis x minus 5 right parenthesis squared end cell less or equal than cell 0 space end cell row cell left parenthesis left parenthesis 2 x minus 1 right parenthesis plus left parenthesis x minus 5 right parenthesis right parenthesis left parenthesis 2 x minus 1 right parenthesis minus left parenthesis x minus 5 right parenthesis end cell less or equal than 0 row cell left parenthesis 2 x minus 1 plus x minus 5 right parenthesis left parenthesis 2 x minus 1 minus x plus 5 right parenthesis end cell less or equal than 0 row cell left parenthesis 3 x minus 6 right parenthesis left parenthesis x plus 4 right parenthesis end cell less or equal than 0 end table end style      

Kita cari batasnya, sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x minus 6 end cell equals cell 0 space atau space x plus 4 equals 0 end cell row cell 3 x end cell equals cell 6 space atau space x equals negative 4 end cell row x equals cell 2 space atau space x equals negative 4 end cell end table  

Kemudian kita uji daerah dari setiap batas melalui garis bilangan sebagai berikut:

Karena pertidaksamaan pada soal bertanda less or equal than 0, maka yang menjadi penyelesaian adalah daerah negatif pada interval negative 4 less or equal than x less or equal than 2.

Dengan demikian, interval nilai x yang memenuhi pertidaksamaan tersebut adalah negative 4 less or equal than x less or equal than 2.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 12 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Tentukan interval nilai  yang memenuhi pertidaksamaan berikut ini : c.

Pembahasan Soal:

Ingat!

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a close vertical bar end cell less than cell open vertical bar b close vertical bar end cell row cell open vertical bar a close vertical bar squared end cell less than cell open vertical bar b close vertical bar squared end cell row cell a squared end cell less than cell b squared end cell row cell a squared minus b squared end cell less than 0 row cell left parenthesis a minus b right parenthesis left parenthesis a plus b right parenthesis end cell less than 0 end table  

Maka:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell vertical line 2 x plus 5 vertical line end cell less than cell vertical line 3 x minus 10 vertical line end cell row cell left parenthesis 2 x plus 5 right parenthesis squared end cell less than cell left parenthesis 3 x minus 10 right parenthesis squared end cell row cell left parenthesis 2 x plus 5 right parenthesis squared minus left parenthesis 3 x minus 10 right parenthesis squared end cell less than 0 row cell left parenthesis left parenthesis 2 x plus 5 right parenthesis plus left parenthesis 3 x minus 10 right parenthesis right parenthesis left parenthesis 2 x plus 5 right parenthesis minus left parenthesis 3 x minus 10 right parenthesis end cell less than 0 row cell left parenthesis 2 x plus 5 plus 3 x minus 10 right parenthesis left parenthesis 2 x plus 5 minus 3 x plus 10 right parenthesis end cell less than 0 row cell left parenthesis 5 x minus 5 right parenthesis left parenthesis negative x plus 15 right parenthesis end cell less than 0 end table end style     

Kita cari batasnya, sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell 5 x minus 5 end cell equals cell 0 space atau space minus x plus 15 equals 0 end cell row cell 5 x end cell equals cell 5 space atau space minus x equals negative 15 end cell row x equals cell 1 space atau space x equals 15 end cell end table 

Kemudian kita uji daerah dari setiap batas melalui garis bilangan sebagai berikut:

Karena pertidaksamaan pada soal bertanda less than 0, maka yang memenuhi adalah interval bertanda negatif atau interval x less than 1 space atau space x greater than 15.

Dengan demikian, interval nilai x yang memenuhi pertidaksamaan tersebut adalah x less than 1 space atau space x greater than 15.

0

Roboguru

Tentukan interval nilai  yang memenuhi pertidaksamaan berikut ini : d.

Pembahasan Soal:

Ingat!

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a close vertical bar end cell greater or equal than cell open vertical bar b close vertical bar end cell row cell open vertical bar a close vertical bar squared end cell greater or equal than cell open vertical bar b close vertical bar squared end cell row cell a squared end cell greater or equal than cell b squared end cell row cell a squared minus b end cell greater or equal than 0 row cell left parenthesis a minus b right parenthesis left parenthesis a plus b right parenthesis end cell greater or equal than 0 end table  

Maka:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell vertical line x plus 4 vertical line end cell greater or equal than cell vertical line 3 x minus 8 vertical line end cell row cell left parenthesis x plus 4 right parenthesis squared end cell greater or equal than cell left parenthesis 3 x minus 8 right parenthesis squared end cell row cell left parenthesis x plus 4 right parenthesis squared minus left parenthesis 3 x minus 8 right parenthesis squared end cell greater or equal than 0 row cell left parenthesis left parenthesis x plus 4 right parenthesis plus left parenthesis 3 x minus 8 right parenthesis right parenthesis left parenthesis left parenthesis x plus 4 right parenthesis minus left parenthesis 3 x minus 8 right parenthesis right parenthesis end cell greater or equal than 0 row cell left parenthesis x plus 4 plus 3 x minus 8 right parenthesis left parenthesis x plus 4 minus 3 x plus 8 right parenthesis end cell greater or equal than 0 row cell left parenthesis 4 x minus 4 right parenthesis left parenthesis negative 2 x plus 12 right parenthesis end cell greater or equal than 0 end table end style 

Kita cari batasnya, sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 x minus 4 end cell equals cell 0 space atau space minus 2 x plus 12 equals 0 end cell row cell 4 x end cell equals cell 4 space atau space minus 2 x equals negative 12 end cell row x equals cell 1 space atau space x equals 6 end cell end table  

Kemudian kita uji daerah dari setiap batas melalui garis bilangan sebagai berikut:

Karena pertidaksamaan pada soal bertanda greater or equal than 0, maka yang memenuhi adalah interval yang bernilai positif atau interval 1 less or equal than x less or equal than 6.

Dengan demikian, interval nilai x yang memenuhi pertidaksamaan tersebut adalah 1 less or equal than x less or equal than 6

0

Roboguru

Penyelesaian pertidaksamaan   adalah …

Pembahasan Soal:

Ingat!

Definisi dari suatu nilai mutlak adalah:

open vertical bar x close vertical bar open curly brackets table row cell x comma space x greater or equal than 0 space end cell row blank row cell negative x comma space x less than 0 end cell end table close 

Maka:

Diketahui x plus open vertical bar x minus 3 close vertical bar less or equal than 3.

open vertical bar x minus 3 close vertical bar open curly brackets table row cell x minus 3 comma space x minus 3 greater or equal than 0 end cell row cell x greater or equal than 3 end cell row cell negative left parenthesis x minus 3 right parenthesis comma space x minus 3 less than 0 end cell row cell x less than 3 end cell end table close 

  • Untuk x greater or equal than 3 

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus x minus 3 end cell less or equal than 3 row cell 2 x minus 3 end cell less or equal than 3 row cell 2 x end cell less or equal than cell 3 plus 3 end cell row cell 2 x end cell less or equal than 6 row x less or equal than cell 6 over 2 end cell row x less or equal than 3 end table 

Untuk interval x greater or equal than 3x less or equal than 3 diriskan dengan x greater or equal than 3 sehingga penyelesaian menjadi x equals 3.

  • Untuk x less than 3 

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus left parenthesis negative left parenthesis x minus 3 right parenthesis right parenthesis end cell less or equal than 3 row cell x minus x plus 3 end cell less or equal than 3 row 3 less or equal than 3 end table 

Untuk interval x less than 3, tidak memiliki penyelesaian.

Sehingga dari 2 interval di atas kita gabungkan menjadi x equals 3.

Dengan demikian, penyelesaian pertidaksamaan tersebut adalah x equals 3.

0

Roboguru

Tentukan interval nilai  yang memenuhi pertidaksamaan berikut ini : a.

Pembahasan Soal:

Ingat!

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a close vertical bar end cell greater than cell open vertical bar b close vertical bar end cell row cell open vertical bar a close vertical bar squared end cell greater than cell open vertical bar b close vertical bar squared end cell row cell a squared end cell greater than cell b squared end cell row cell a squared minus b squared end cell greater than 0 row cell left parenthesis a minus b right parenthesis left parenthesis a plus b right parenthesis end cell greater than 0 end table  

Maka:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 4 x minus 1 close vertical bar end cell greater than cell open vertical bar 3 x minus 6 close vertical bar end cell row cell left parenthesis 4 x minus 1 right parenthesis squared end cell greater than cell open parentheses 3 x minus 6 close parentheses squared space end cell row cell left parenthesis 4 x minus 1 right parenthesis squared minus open parentheses 3 x minus 6 close parentheses squared end cell greater than 0 row cell left parenthesis left parenthesis 4 x minus 1 right parenthesis plus left parenthesis 3 x minus 6 right parenthesis right parenthesis left parenthesis 4 x minus 1 right parenthesis minus left parenthesis 3 x minus 6 right parenthesis end cell greater than 0 row cell left parenthesis 4 x minus 1 plus 3 x minus 6 right parenthesis left parenthesis 4 x minus 1 minus 3 x plus 6 right parenthesis end cell greater than 0 row cell left parenthesis 7 x minus 7 right parenthesis left parenthesis x plus 5 right parenthesis end cell greater than 0 end table end style   

Kita cari batasnya, sehingga:

7 x minus 7 equals 0 space atau space x plus 5 equals 0 7 x equals 7 space atau space x equals negative 5 x equals 1 space atau space x equals negative 5 

Kemudian kita uji daerah dari setiap batas melalui garis bilangan sebagai berikut:

Karena pertidaksamaan bertanda greater than 0, maka yang diambil adalah interval yang positif yaitu x less than negative 5 space atau space x greater than 1.

Dengan demikian, interval nilai x yang memenuhi pertidaksamaan tersebut adalah x less than negative 5 space atau space x greater than 1.

0

Roboguru

Nilai-nilai yang memenuhi  adalah...

Pembahasan Soal:

Perhatikan penghitungan berikut!

 Sifat space nilai space mutlak open vertical bar f open parentheses x close parentheses close vertical bar less or equal than open vertical bar g open parentheses x close parentheses close vertical bar left right arrow f open parentheses x close parentheses squared less or equal than g open parentheses x close parentheses squared  open vertical bar x plus 3 close vertical bar less or equal than open vertical bar 2 x close vertical bar open parentheses x plus 3 close parentheses squared less or equal than open parentheses 2 x close parentheses squared open parentheses x squared plus 6 x plus 9 close parentheses less or equal than 4 x squared x squared plus 6 x plus 9 minus 4 x squared less or equal than 0 minus 3 x squared plus 6 x plus 9 less or equal than 0 x squared minus 2 x minus 3 greater or equal than 0 left parenthesis x minus 3 right parenthesis left parenthesis x plus 1 right parenthesis greater or equal than 0 Uji space titik colon x minus 3 equals 0 rightwards arrow x equals 3 x plus 1 equals 0 rightwards arrow x equals negative 1  untuk space x less or equal than negative 1 misal space x equals negative 2 rightwards arrow open parentheses negative 2 minus 3 close parentheses open parentheses negative 2 plus 1 close parentheses equals 5 space open parentheses positif close parentheses  untuk space minus 1 less or equal than x less or equal than 3 misal space x equals 0 rightwards arrow open parentheses 0 minus 3 close parentheses open parentheses 0 plus 1 close parentheses equals negative 3 space open parentheses negatif close parentheses  untuk space x greater or equal than 3 misal space x equals 4 rightwards arrow open parentheses 4 minus 3 close parentheses open parentheses 4 plus 1 close parentheses equals 5 space open parentheses positif close parentheses             

Jadi, HP equals open curly brackets x less or equal than negative 1 space atau space x greater or equal than 3 space close curly brackets. space   

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved