Roboguru
SD
SMP
SMA
UTBK/SNBT

Iklan

Pertanyaan

Himpunan penyelesaian dari pertidaksamaan 9 ( p + 1 ) ( p + 3 ) > ( 2 7 p ) p ⋅ 8 1 2 p adalah ....

Himpunan penyelesaian dari pertidaksamaan   adalah ....

  1. begin mathsize 14px style open curly brackets straight p │ minus square root of 6 less than straight p less than square root of 6 comma straight p element of straight R close curly brackets end style 

  2. begin mathsize 14px style open curly brackets straight p │ straight p less than negative square root of 6 space ataup greater than square root of 6 comma straight p element of straight R close curly brackets end style 

  3. begin mathsize 14px style open curly brackets straight p │ minus 6 less than straight p less than 6 comma straight p element of straight R close curly brackets end style 

  4. begin mathsize 14px style open curly brackets straight p │ straight p less than negative 6 space atau space straight p greater than 6 comma straight p element of straight R close curly brackets end style 

  5. begin mathsize 14px style open curly brackets straight p │ straight p element of straight R close curly brackets end style 

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

00

:

15

:

55

:

52

Klaim

Iklan

M. Robo

Master Teacher

Jawaban terverifikasi

Pembahasan

Perhatikan bahwa Karena 3 > 1, maka Didapat pembuat nol yaitu atau Sehingga didapat penyelesaian . Maka, himpunan penyelesaian dari pertidaksamaan adalah

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 9 to the power of left parenthesis straight p plus 1 right parenthesis left parenthesis straight p plus 3 right parenthesis end exponent end cell greater than cell left parenthesis 27 to the power of straight p right parenthesis to the power of straight p times 81 squared straight p end cell row cell 9 to the power of straight p squared plus 4 straight p plus 3 end exponent end cell greater than cell left parenthesis 27 to the power of straight p right parenthesis to the power of straight p times 81 squared straight p end cell row cell left parenthesis 3 squared right parenthesis to the power of straight p squared plus 4 straight p plus 3 end exponent end cell greater than cell left parenthesis left parenthesis 3 cubed right parenthesis to the power of straight p right parenthesis to the power of straight p times left parenthesis 3 to the power of 4 right parenthesis squared straight p end cell row cell 3 to the power of 2 straight p squared plus 8 straight p plus 6 end exponent end cell greater than cell 3 to the power of 3 straight p squared end exponent times 3 to the power of 8 straight p end cell row cell 3 to the power of 2 straight p squared plus 8 straight p plus 6 end exponent end cell greater than cell 3 to the power of 3 straight p squared plus 8 straight p end exponent end cell row blank blank blank end table end style 

Karena 3 > 1, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 straight p squared plus 8 straight p plus 6 end cell greater than cell 3 straight p squared plus 8 straight p end cell row cell negative straight p squared plus 6 end cell greater than 0 row cell straight p squared minus 6 end cell less than 0 row cell left parenthesis straight p plus square root of 6 right parenthesis left parenthesis straight p minus square root of 6 right parenthesis end cell less than 0 end table end style 

Didapat pembuat nol yaitu begin mathsize 14px style space straight p equals negative square root of 6 end style atau begin mathsize 14px style straight p equals square root of 6 end style 

Sehingga didapat penyelesaian begin mathsize 14px style space minus square root of 6 less than straight p less than square root of 6 end style  .

Maka, himpunan penyelesaian dari pertidaksamaan begin mathsize 14px style 9 to the power of left parenthesis straight p plus 1 right parenthesis left parenthesis straight p plus 3 right parenthesis end exponent greater than left parenthesis 27 to the power of straight p right parenthesis to the power of straight p times 81 squared straight p end style adalah undefined 

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

1

Iklan

Gold Icon

Klaim Gold gratis sekarang!

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

Pertanyaan serupa

Banyaknya bilangan bulat p yang memenuhi pertidaksamaan 6 4 p 2 − 2 p + 3 ≥ ( 3 2 p ⋅ 8 p − 1 ) p ada sebanyak ... buah.

1

5.0

Jawaban terverifikasi

Himpunan penyelesaian dari pertidaksamaan 4 p 2 + p + 12 > ( 8 p ) p ⋅ 1 6 p adalah ....

1

0.0

Jawaban terverifikasi

Iklan

Bilangan bulat terbesar yang memenuhi pertidaksamaan adalah ....

1

5.0

Jawaban terverifikasi

Bilangan bulat terbesar yang memenuhi pertidaksamaan adalah ....

2

5.0

Jawaban terverifikasi

Himpunan penyelesaian dari pertidaksamaan 5 p 2 − 2 p + 4 < ( 2 5 p ) p ⋅ 12 5 2 p adalah ....

2

0.0

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 02130930000

02130930000

Ikuti Kami

©2025 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia