Roboguru

Himpunan penyelesaian dari pertidaksamaan  adalah ….

Pertanyaan

Himpunan penyelesaian dari pertidaksamaan begin mathsize 14px style open vertical bar 2 x minus 1 close vertical bar greater than vertical line 4 x plus 3 vertical line end style adalah ….space   

  1. begin mathsize 14px style open curly brackets negative 2 less than x less than negative 1 third close curly brackets end style     undefined 

  2. begin mathsize 14px style open curly brackets 1 third less than x less than 2 close curly brackets end style     undefined 

  3. begin mathsize 14px style open curly brackets negative 2 less than x less than 1 third close curly brackets end style     undefined 

  4. begin mathsize 14px style open curly brackets negative 2 less or equal than x less or equal than 1 third close curly brackets end style      begin mathsize 14px style space end style 

  5. begin mathsize 14px style open curly brackets 1 third less or equal than x less than 2 close curly brackets end style     undefined 

Pembahasan Soal:

Berdasarkan sifat undefined 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x minus 1 close vertical bar end cell greater than cell vertical line 4 x plus 3 vertical line end cell row cell square root of open parentheses 2 x minus 1 close parentheses squared end root end cell greater than cell square root of open parentheses 4 x plus 3 close parentheses squared end root end cell row blank blank blank end table end style 

Kuadratkan kedua ruas

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses square root of open parentheses 2 x minus 1 close parentheses squared end root close parentheses squared end cell greater than cell open parentheses square root of open parentheses 4 x plus 3 close parentheses squared end root close parentheses squared end cell row cell 4 x squared minus 4 x plus 1 end cell greater than cell 16 x squared plus 24 x plus 9 end cell row cell 4 x squared minus 16 x squared minus 4 x minus 24 x plus 1 minus 9 end cell greater than 0 row cell negative 12 x squared minus 28 x minus 8 end cell greater than 0 row cell 12 x squared plus 28 x plus 8 end cell less than 0 row blank blank blank row blank blank blank row blank blank blank row blank blank blank end table end style 

Lakukan pemfaktoran

begin mathsize 14px style open parentheses 6 x plus 2 close parentheses open parentheses 2 x plus 4 close parentheses less than 0 end style 

Menentukan pembuat nol

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 6 x plus 2 end cell equals 0 row cell 6 x end cell equals cell negative 2 end cell row x equals cell negative 1 third end cell row blank blank blank row blank blank blank end table end style 

atau

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x plus 4 end cell equals 0 row cell 2 x end cell equals cell negative 4 end cell row x equals cell negative 2 end cell end table end style 

Maka diperoleh interval sebagai berikut

begin mathsize 14px style HP equals left curly bracket x vertical line minus 2 less than x less than negative 1 third right curly bracket end style 

Jadi, himpunan penyelesaian dari pertidaksamaan begin mathsize 14px style begin bold style vertical line 2 x minus 1 vertical line end style bold greater than bold vertical line bold 4 bold italic x bold plus bold 3 bold vertical line end style adalah A. begin mathsize 14px style begin bold style left curly bracket negative 2 less than x less than negative 1 third right curly bracket end style end style

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Alfi

Mahasiswa/Alumni Universitas Negeri Semarang

Terakhir diupdate 05 Oktober 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 x yang memenuhi pertidaksamaan berikut: 8. ∣x+1∣>∣3x−5∣

Pembahasan Soal:

Ingat, salah satu cara menyelesaikan permasalahan pertidaksamaan nilai mutlak adalah dengan mengkuadratkan kedua ruas:

open vertical bar f left parenthesis x right parenthesis close vertical bar greater than open vertical bar g left parenthesis x right parenthesis close vertical bar left right arrow open parentheses f left parenthesis x right parenthesis close parentheses squared greater than open parentheses g left parenthesis x right parenthesis close parentheses squared

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x plus 1 close vertical bar end cell greater than cell open vertical bar 3 x minus 5 close vertical bar end cell row cell left parenthesis x plus 1 right parenthesis squared end cell greater than cell left parenthesis 3 x minus 5 right parenthesis squared space end cell row cell x squared plus 2 x plus 1 end cell greater than cell 9 x squared minus 30 x plus 25 end cell row cell 8 x squared minus 32 x plus 24 end cell less than 0 row cell x squared minus 4 x plus 3 end cell less than 0 row cell left parenthesis x minus 3 right parenthesis left parenthesis x minus 1 right parenthesis end cell less than 0 end table

Dengan menggunakan uji titik 0, maka didapatkan interval penyelesaiannya adalah 1 less than x less than 3.

Jadi, interval nilai x yang memenuhi adalah 1 less than x less than 3.

0

Roboguru

Himpunan penyelesaian dari pertidaksamaan ∣2x+1∣≥∣x−2∣ adalah...

Pembahasan Soal:

  • Menyelesaikan pertidaksamaan nilai mutlak:

open vertical bar f left parenthesis x right parenthesis close vertical bar greater or equal than open vertical bar g left parenthesis x right parenthesis close vertical bar left right double arrow open square brackets f left parenthesis x right parenthesis close square brackets squared greater or equal than open square brackets g left parenthesis x right parenthesis close square brackets squared

sedemikian hingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x plus 1 close vertical bar end cell greater or equal than cell open vertical bar x minus 2 close vertical bar end cell row cell open square brackets 2 x plus 1 close square brackets squared end cell greater or equal than cell open square brackets x minus 2 close square brackets squared end cell row cell 4 x squared plus 4 x plus 1 end cell greater or equal than cell x squared minus 4 x plus 4 end cell row cell 3 x squared plus 8 x minus 3 end cell greater or equal than 0 end table

Pembuat nol:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x squared plus 8 x minus 3 end cell equals 0 row cell open parentheses x plus 3 close parentheses open parentheses 3 x minus 1 close parentheses end cell equals 0 end table

x equals negative 3 space atau space x equals 1 third

Garis bilangan:

Jadi, himpunan penyelesaian dari pertidaksamaan tersebut adalah open curly brackets x vertical line x less or equal than negative 3 space atau space x greater or equal than 1 third comma space x element of R close curly brackets.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Penyelesaian pertidaksamaan ∣x+4∣≥∣2x+1∣ adalah ...

Pembahasan Soal:

Ingat bahwa:

Nilai mutlak memiliki sifat open vertical bar a close vertical bar squared equals a squared. Selain itu, kita ketahui juga bahwa bentuk open parentheses a plus b close parentheses squared equals a squared plus 2 a b plus b squared.

Sehingga berdasarkan dua sifat di atas diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x plus 4 close vertical bar end cell greater or equal than cell open vertical bar 2 x plus 1 close vertical bar end cell row cell open parentheses x plus 4 close parentheses squared end cell greater or equal than cell open parentheses 2 x plus 1 close parentheses squared end cell row cell x squared plus 8 x plus 16 end cell greater or equal than cell 4 x squared plus 4 x plus 1 end cell row cell x squared minus 4 x squared plus 8 x minus 4 x plus 16 minus 1 end cell greater or equal than 0 row cell negative 3 x squared plus 4 x plus 15 end cell greater or equal than 0 row cell 3 x squared minus 4 x minus 15 end cell less or equal than 0 end table

Untuk mencari nilai x ubahlah pertidaksamaan di atas menjadi persamaan dan faktorkanlah persamaan tersebut.

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x squared minus 4 x minus 15 end cell less or equal than 0 row cell 3 x squared minus 4 x minus 15 end cell equals 0 row cell open parentheses 3 x plus 5 close parentheses open parentheses x minus 3 close parentheses end cell equals 0 row cell 3 x plus 5 end cell equals 0 row cell 3 x end cell equals cell negative 5 end cell row x equals cell negative 5 over 3 end cell row x equals cell negative 1 2 over 3 end cell row blank blank atau row cell x minus 3 end cell equals 0 row x equals 3 end table

Sehingga diperoleh x equals negative 1 2 over 3 space atau space x equals 3.

Kemudian ambil sembarang titik x greater or equal than 3 comma space minus 1 2 over 3 less than x less than 3 comma space dan space x less or equal than negative 1 2 over 3 dan substitusikan pada 3 x squared minus 4 x minus 15 less or equal than 0.

Misal 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 equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 open parentheses 4 close parentheses squared minus 4 open parentheses 4 close parentheses minus 15 end cell less or equal than 0 row cell 3 open parentheses 16 close parentheses minus 16 minus 15 end cell less or equal than 0 row cell 48 minus 31 end cell less or equal than 0 row 17 less or equal than cell 0 horizontal ellipsis left parenthesis salah right parenthesis end cell end table

Misal x equals 0

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 open parentheses 0 close parentheses squared minus 4 open parentheses 0 close parentheses minus 15 end cell less or equal than 0 row cell 0 minus 0 minus 15 end cell less or equal than 0 row cell negative 15 end cell less or equal than cell 0 horizontal ellipsis open parentheses benar close parentheses end cell end table

Misal x equals negative 2

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 open parentheses negative 2 close parentheses squared minus 4 open parentheses negative 2 close parentheses minus 15 end cell less or equal than 0 row cell 3 open parentheses 4 close parentheses plus 8 minus 15 end cell less or equal than 0 row cell 12 minus 7 end cell less or equal than 0 row 5 less or equal than cell 0 horizontal ellipsis open parentheses salah close parentheses end cell end table

Berdasarkan pemisalan di atas diperoleh daerah penyelesaian seperti pada garis bilangan berikut.

Sehingga daerah penyelesaian dari pertidaksamaan open vertical bar x plus 4 close vertical bar greater or equal than open vertical bar 2 x plus 1 close vertical bar adalah negative 1 2 over 3 less or equal than x less or equal than 3.

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Penyelesaian pertidaksamaan ∣x+2∣≥∣x−3∣ adalah...

Pembahasan Soal:

Ingat kembali sifat berikut!

left parenthesis a plus b right parenthesis squared equals a to the power of 2 end exponent plus 2 a b plus b squared left parenthesis a minus b right parenthesis squared equals a squared minus 2 a b plus b squared 

Dari sifat di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x plus 2 close vertical bar end cell greater or equal than cell open vertical bar x minus 3 close vertical bar end cell row cell open vertical bar x plus 2 close vertical bar squared end cell greater or equal than cell open vertical bar x minus 3 close vertical bar squared end cell row cell left parenthesis x plus 2 right parenthesis squared end cell greater or equal than cell left parenthesis x minus 3 right parenthesis squared space end cell row cell x squared plus 2 times x times 2 plus 2 squared end cell greater or equal than cell x squared minus 2 times x times 3 plus 3 squared end cell row cell x squared plus 4 x plus 4 end cell greater or equal than cell x squared minus 6 x plus 9 end cell row cell x squared minus x squared plus 4 x plus 6 x plus 4 minus 9 end cell greater or equal than 0 row cell 10 x minus 5 end cell greater or equal than 0 row cell 10 x end cell greater or equal than 5 row x greater or equal than cell 5 over 10 end cell row x greater or equal than cell 0 comma 5 end cell end table   

​​​​​​​Oleh karena itu, jawaban yang benar adalah A. 

1

Roboguru

Penyelesaian pertidaksamaan ∣2x+4∣≥∣x+5∣ adalah ....

Pembahasan Soal:

open vertical bar 2 x plus 4 close vertical bar greater or equal than open vertical bar x plus 5 close vertical bar open parentheses 2 x plus 4 close parentheses squared greater or equal than open parentheses x plus 5 close parentheses squared open parentheses 2 x plus 4 close parentheses squared minus open parentheses x plus 5 close parentheses squared greater or equal than 0 open parentheses 2 x plus 4 plus x plus 5 close parentheses open parentheses 2 x plus 4 minus x minus 5 close parentheses greater or equal than 0 open parentheses 3 x plus 9 close parentheses open parentheses x minus 1 close parentheses greater or equal than 0 

Kita cari pembuat nol fungsi dulu

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x plus 9 end cell equals cell 0 space logical or space x minus 1 equals 0 end cell row cell 3 x end cell equals cell negative 9 space space space space space space space space space space x equals 1 end cell row x equals cell negative 3 end cell end table 

Buat garis bilangan

x less or equal than negative 3 space logical or x greater or equal than 1 

Jadi, Penyelesaian pertidaksamaan open vertical bar 2 x plus 4 close vertical bar greater or equal than open vertical bar x plus 5 close vertical bar adalah x less or equal than negative 3 space logical or x greater or equal than 1.

 

 

3

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