Roboguru
SD
SMP
SMA
UTBK/SNBT

Iklan

Iklan

Pertanyaan

Diketahui pertidaksamaan x 2 − ∣ x ∣ ≤ 12 . Himpunan penyelesaian pertidaksamaan tersebut adalah …

Diketahui pertidaksamaan  . Himpunan penyelesaian pertidaksamaan tersebut adalah …

  1. open curly brackets negative 4 less or equal than x less than 0 close curly brackets

  2. open curly brackets 0 less or equal than x less than 4 close curly brackets

  3. open curly brackets negative 4 less or equal than x less than 3 close curly brackets

  4. open curly brackets negative 3 less or equal than x less than 4 close curly brackets

  5. open curly brackets negative 4 less or equal than x less or equal than 4 close curly brackets

Ke roboguru plus

Iklan

N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Iklan

Pembahasan

x squared minus open vertical bar x close vertical bar less or equal than 12 x squared minus open vertical bar x close vertical bar minus 12 less or equal than 0 K a r e n a space t e r d a p a t space n i l a i space m u t l a k comma space m a k a space a k a n space a d a space d u a space k e m u n g k i n a n comma space y a i t u colon 1. space J i k a space x less than 0 comma space m a k a space open vertical bar x close vertical bar equals negative x x squared minus open vertical bar x close vertical bar minus 12 less or equal than 0 x squared minus open parentheses negative x close parentheses minus 12 less or equal than 0 x squared plus x minus 12 less or equal than 0 open parentheses x plus 4 close parentheses open parentheses x minus 3 close parentheses less or equal than 0 x subscript 1 equals negative 4 space a t a u space x subscript 2 equals 3 U j i space d a n space g a m b a r space g a r i s space b i l a n g a n n y a

M a k a space i r i s a n space a n t a r a space x less than 0 space d a n minus 4 less or equal than x less or equal than 3

S e h i n g g a space h i m p u n a n space p e n y e l e s a i a n space u n t u k space k e m u n g k i n a n space I equals open curly brackets negative 4 less or equal than x less than 0 close curly brackets 2. space J i k a space x greater or equal than 0 comma space m a k a space open vertical bar x close vertical bar equals x x squared minus x minus 12 less or equal than 0 open parentheses x minus 4 close parentheses open parentheses x plus 3 close parentheses less or equal than 0 x subscript 1 equals 4 space a t a u space x subscript 2 equals negative 3 U j i space d a n space g a m b a r space g a r i s space b i l a n g a n n y a

M a k a space i r i s a n space a n t a r a space x greater or equal than 0 space d a n space minus 3 less or equal than x less or equal than 4 comma y a i t u colon

S e h i n g g a space h i m p u n a n space p e n y e l e s a i a n space u n t u k k e m u n g k i n a n space I I comma space y a i t u space equals open curly brackets 0 less or equal than x less or equal than 4 close curly brackets J i k a space d i g a b u n g k a n space a n t a r a space k e m u n g k i n a n space I d a n space k e m u n g k i n a n space I I comma space m a k a colon K e m u n g k i n a n space I equals open curly brackets negative 4 less or equal than x less than 0 close curly brackets K e m u n g k i n a n space I I equals open curly brackets 0 less or equal than x less or equal than 4 close curly brackets J a d i comma space h i m p u n a n space p e n y e l e s a i a n n y a a d a l a h colon space open curly brackets negative 4 less or equal than x less than 0 space a t a u space 0 less than x less or equal than 4 close curly brackets

 

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

34

Iklan

Iklan

Gold Icon

Klaim Gold gratis sekarang!

Dengan Gold kamu bisa tanya soal ke Forum sepuasnya, lho.

Pertanyaan serupa

Rogu pergi ke toko alat tulis untuk membeli sebuah tas baru. Harga sebuah tas di toko tersebut adalah 12 kali harga sebuah buku tulis. Harga 3 buah tas dan 6 buah buku tulis adalah Rp302.400,00. Harga...

17

0.0

Jawaban terverifikasi

Nilai x yang memenuhi persamaan ∣ x − 3 ∣ + ∣ 2 x − 8 ∣ = 5 adalah...

109

0.0

Jawaban terverifikasi

Iklan

Iklan

Himpunan penyelesaian dari pertidaksamaan 2 ( x + 3 ) ≤ 3 ( 2 x − 1 ) adalah …

23

0.0

Jawaban terverifikasi

Persamaan 2 x 3 ​ = x − 2 1 ​ mempunyai penyelesaian x = . . .

1

5.0

Jawaban terverifikasi

Penyelesaian pertidaksamaan ∣ x − 3∣ ≤ 1 adalah . . .

94

5.0

Jawaban terverifikasi
Ruangguru

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Download di Google Play

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Produk Lainnya

Bantuan & Panduan

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