Iklan

Iklan

Pertanyaan

Pertidaksamaan ∣ ∣ ​ x − 1 2 x + 7 ​ ∣ ∣ ​ ≥ 1 dipenuhi oleh x dalam interval ...

Pertidaksamaan  dipenuhi oleh  dalam interval ...

  1. open square brackets negative 2 comma space 8 close square brackets

  2. left parenthesis negative infinity comma space minus 8 right square bracket union left square bracket negative 2 comma space infinity right parenthesis 

  3. left square bracket negative 8 comma space 1 right parenthesis union open parentheses 1 comma space infinity close parentheses

  4. left square bracket negative 2 comma space 1 right parenthesis union left parenthesis 1 comma space 8 right square bracket

  5. open parentheses infinity comma space minus 8 close parentheses union left square bracket negative 2 comma space 1 right parenthesis union left parenthesis 1 comma space infinity right parenthesis

Iklan

I. Sutiawan

Master Teacher

Mahasiswa/Alumni Universitas Pasundan

Jawaban terverifikasi

Iklan

Pembahasan

Sehingga penyelesaian pertidaksamaan adalah atau dapat ditulis dalam bentuk . Jadi, jawaban yang tepat adalah B

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator 2 x plus 7 over denominator x minus 1 end fraction close vertical bar end cell greater or equal than 1 row cell fraction numerator open vertical bar 2 x plus 7 close vertical bar over denominator open vertical bar x minus 1 close vertical bar end fraction end cell greater or equal than 1 row cell open vertical bar 2 x plus 7 close vertical bar end cell greater or equal than cell open vertical bar x minus 1 close vertical bar end cell row cell left parenthesis left parenthesis 2 x plus 7 right parenthesis plus left parenthesis x minus 1 right parenthesis right parenthesis left parenthesis 2 x plus 7 right parenthesis minus left parenthesis x minus 1 right parenthesis right parenthesis end cell greater or equal than 0 row cell left parenthesis 2 x plus 7 plus x minus 1 right parenthesis left parenthesis 2 x plus 7 minus x plus 1 right parenthesis end cell greater or equal than 0 row cell left parenthesis 3 x plus 6 right parenthesis left parenthesis x plus 8 right parenthesis end cell greater or equal than 0 end table

                             table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank less or equal than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 8 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank atau end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank greater or equal than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table

Sehingga penyelesaian pertidaksamaan open vertical bar fraction numerator 2 x plus 7 over denominator x minus 1 end fraction close vertical bar greater or equal than 1 adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank less or equal than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 8 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank atau end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank greater or equal than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table atau dapat ditulis dalam bentuk left parenthesis negative infinity comma space minus 8 right square bracket space union left square bracket negative 2 comma space infinity right parenthesis.

Jadi, jawaban yang tepat adalah B

 

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

2

Davy Anhar

Bantu banget Ini yang aku cari!

Iklan

Iklan

Pertanyaan serupa

Tentukan himpunan penyelesaian dari setiap pertidaksamaan nilai mutlak di bawah ini. d. ∣ ∣ ​ x 2 − 5 ∣ ∣ ​ ≥ 4

131

0.0

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia