Roboguru

Pertanyaan

Nilai x yang memenuhi pertidaksamaan fraction numerator x squared minus 2 x minus 12 over denominator x minus 2 end fraction less than 1 adalah ...

  1. negative 2 less than x less than 5

  2. negative 5 less than x less than 2

  3. x less than negative 2 blank atau blank x greater than 5

  4. x less than negative 5 blank atau blank x greater than 2

  5. negative 3 less than x less than 10

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar tidak ada.

Pembahasan

Ingat cara menyelesaikan dan menentukan himpunan penyelesaian dari suatu pertidaksamaan.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x squared minus 2 x minus 12 over denominator x minus 2 end fraction end cell less than 1 row cell fraction numerator x squared minus 2 x minus 12 over denominator x minus 2 end fraction minus 1 end cell less than 0 row cell fraction numerator x squared minus 2 x minus 12 over denominator x minus 2 end fraction minus fraction numerator x minus 2 over denominator x minus 2 end fraction end cell less than 0 row cell fraction numerator x squared minus 2 x minus 12 minus left parenthesis x minus 2 right parenthesis over denominator x minus 2 end fraction end cell less than 0 row cell fraction numerator x squared minus 2 x minus 12 minus x plus 2 over denominator x minus 2 end fraction end cell less than 0 row cell fraction numerator x squared minus 2 x minus x minus 12 plus 2 over denominator x minus 2 end fraction end cell less than 0 row cell fraction numerator x squared minus 3 x minus 10 over denominator x minus 2 end fraction end cell less than 0 end table

Didapatkan fraction numerator x squared minus 3 x minus 10 over denominator x minus 2 end fraction less than 0, kasus I dimana x squared minus 3 x minus 10 less than 0 dan x minus 2 greater than 0.

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 3 x minus 10 end cell less than 0 row cell left parenthesis x minus 5 right parenthesis left parenthesis x plus 2 right parenthesis end cell less than 0 row cell x minus 5 end cell less than cell 0 blank atau blank x plus 2 less than 0 end cell row x less than cell 5 blank atau blank x greater than negative 2 end cell end table

dan

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 2 end cell greater than 0 row x greater than 2 end table

Himpunan penyelesainnya adalah x vertical line 2 less than x less than 5.

Kasus II dimana x squared minus 3 x minus 10 greater than 0 dan blank x minus 2 less than 0.

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 3 x minus 10 end cell greater than 0 row cell open parentheses x minus 5 close parentheses open parentheses x plus 2 close parentheses end cell greater than 0 row cell x minus 5 end cell greater than cell 0 blank atau blank x plus 2 greater than 0 end cell row x greater than cell 5 blank atau blank x less than negative 2 end cell end table

dan 

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 2 end cell less than 0 row x less than 2 end table

 Penyelesainnya adalah x less than negative 2 blank atau blank x greater than 5.

Jadi, Penyelesaiannya adalah x less than negative 2 blank atau blank 2 less than x less than 5 blank atau blank x greater than 5.

Oleh karena itu, jawaban yang benar tidak ada.

21

0.0 (0 rating)

Pertanyaan serupa

Tentukan nilai x yang memenuhi pertidaksamaan kuadrat berikut. 16x2>9

8

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 Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia