Iklan

Iklan

Pertanyaan

Himpunan penyelesaian dari pertidaksamaan ∣ ∣ ​ x 2 + 2 x − 9 ∣ ∣ ​ ≤ 6 adalah ...

Himpunan penyelesaian dari pertidaksamaan  adalah ...

  1. open curly brackets negative 5 less or equal than x less or equal than negative 1 space atau space 1 less or equal than x less or equal than 5 close curly brackets

  2. open curly brackets negative 5 less or equal than x less or equal than negative 3 space atau space minus 1 less or equal than x less or equal than 3 close curly brackets

  3. open curly brackets negative 5 less or equal than x less or equal than negative 3 space atau space 1 less or equal than x less or equal than 3 close curly brackets

  4. open curly brackets negative 3 less or equal than x less or equal than negative 1 space atau space 1 less or equal than x less or equal than 5 close curly brackets

  5. open curly brackets negative 3 less or equal than x less or equal than 1 space atau space 3 less or equal than x less or equal than 5 close curly brackets

Iklan

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C

jawaban yang tepat adalah C

Iklan

Pembahasan

Pembahasan
lock

Kita pecah menjadi: . Sehingga: dan Iriskan (1) dan (2), sehingga daerah penyelesaian menjadi . Jadi, jawaban yang tepat adalah C

open vertical bar x squared plus 2 x minus 9 close vertical bar less or equal than 6 minus 6 less or equal than x squared plus 2 x minus 9 less or equal than 6 

Kita pecah menjadi: negative 6 less or equal than x squared plus 2 x minus 9 space dan space x squared plus 2 x minus 9 less or equal than 6. Sehingga:

Error converting from MathML to accessible text.
table attributes columnalign right center left columnspacing 0px end attributes row cell negative x squared minus 2 x plus 9 minus 6 end cell less or equal than 0 row cell negative x squared minus 2 x plus 3 end cell less or equal than 0 row cell x squared plus 2 x minus 3 end cell greater or equal than 0 row cell left parenthesis x plus 3 right parenthesis left parenthesis x minus 1 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 3 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 1 end table space.... space left parenthesis 1 right parenthesis

dan

 table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus 2 x minus 9 end cell less or equal than 6 row cell x squared plus 2 x minus 9 minus 6 end cell less or equal than 0 row cell x squared plus 2 x minus 15 end cell less or equal than 0 row cell left parenthesis x plus 5 right parenthesis left parenthesis x minus 3 right parenthesis end cell less or equal than 0 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 5 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 x less or equal than cell 3 space.... space left parenthesis 2 right parenthesis end cell end table 

Iriskan (1) dan (2), sehingga daerah penyelesaian menjadi negative 5 less or equal than x less or equal than negative 3 space atau space 1 less or equal than x less or equal than 3

Jadi, jawaban yang tepat adalah C

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

2

Iklan

Iklan

Pertanyaan serupa

Nilai x yang memenuhi pertidaksamaan ∣ ∣ ​ x 2 − 9 ∣ ∣ ​ − 7 ≤ 0 adalah ....

6

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