Roboguru

Nilai  dari persamaan nilai mutlak  adalah

Pertanyaan

Nilai x dari persamaan nilai mutlak open vertical bar fraction numerator 2 x minus 1 over denominator 4 x plus 3 end fraction close vertical bar equals 1 adalah ....

  1. negative 3 text  atau end text space minus 2

  2. negative 2 space text atau end text space 3

  3. negative 1 space text atau end text space 1

  4. negative 2 space text atau end text space minus 1

Pembahasan Soal:

Diketahui persamaan nilai mutlak open vertical bar fraction numerator 2 x minus 1 over denominator 4 x plus 3 end fraction close vertical bar equals 1. Berdasarkan sifat persamaan mutlak open vertical bar y close vertical bar equals k yaitu open parentheses open vertical bar y close vertical bar close parentheses squared equals k squared maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator 2 x minus 1 over denominator 4 x plus 3 end fraction close vertical bar end cell equals 1 row cell open parentheses fraction numerator 2 x minus 1 over denominator 4 x plus 3 end fraction close parentheses squared end cell equals cell 1 squared end cell row cell fraction numerator 4 x squared minus 4 x plus 1 over denominator 16 x squared plus 24 x plus 9 end fraction end cell equals 1 row cell 4 x squared minus 4 x plus 1 end cell equals cell 16 x squared plus 24 x plus 9 end cell row cell 16 x squared minus 4 x squared plus 24 x plus 4 x plus 9 minus 1 end cell equals 0 row cell 12 x squared plus 28 x plus 8 end cell equals 0 row cell 3 x squared plus 7 x plus 2 end cell equals 0 row cell open parentheses 3 x plus 1 close parentheses open parentheses x plus 2 close parentheses end cell equals 0 row blank blank cell table attributes columnalign left center center end attributes row cell table attributes columnalign left end attributes row cell table attributes columnalign left end attributes row cell table row cell 3 x plus 1 equals 0 end cell end table end cell end table end cell end table end cell logical or cell x equals negative 2 end cell row cell space space space space space space space 3 x equals negative 1 end cell blank blank row cell space space space space space space space space space x equals fraction numerator negative 1 over denominator 3 end fraction end cell blank blank end table end cell end table

Jadi nilai x dari persamaan nilai mutlak open vertical bar fraction numerator 2 x minus 1 over denominator 4 x plus 3 end fraction close vertical bar equals 1 adalah negative 2 space text atau end text space minus 1 third.

Oleh karena itu, tidak ada jawaban yang benar. Jawaban yang benar adalah negative 2 space text atau end text space minus 1 third.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Febrianti

Mahasiswa/Alumni Universitas Negeri Malang

Terakhir diupdate 13 Agustus 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

Himpunan bilangan real  yang merupakan penyelesaian dari persamaan  adalah ....

Pembahasan Soal:

Berdasarkan definisi nilai mutlak, diperoleh nilai-nilai sebagai berikut.

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction close vertical bar end cell equals cell 3 open curly brackets table row cell fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction equals 3 comma end cell cell jika blank fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction greater or equal than 0 end cell row cell fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction equals negative 3 comma end cell cell jika blank fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction less than 0 end cell end table close end cell end table end style

Selanjutnya, cari masing-masing penyelesaian dari bentuk di atas.

Perhatikan perhitungan berikut!

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction end cell equals cell 3 semicolon space x not equal to negative 1 end cell row cell fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction times open parentheses x plus 1 close parentheses end cell equals cell 3 times open parentheses x plus 1 close parentheses end cell row cell x squared plus 2 x minus 1 end cell equals cell 3 x plus 3 end cell row cell x squared minus x end cell equals 4 row cell open parentheses x minus 1 half close parentheses squared minus 1 fourth end cell equals 4 row cell open parentheses x minus 1 half close parentheses squared end cell equals cell 4 plus 1 fourth end cell row cell open parentheses x minus 1 half close parentheses squared end cell equals cell 17 over 4 end cell row cell x minus 1 half end cell equals cell plus-or-minus square root of 17 over 4 end root end cell row x equals cell 1 half plus-or-minus 1 half square root of 17 end cell row x equals cell 1 half open parentheses 1 plus-or-minus square root of 17 close parentheses end cell end table

Dari persamaan begin mathsize 14px style fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction equals 3 end style, diperoleh penyelesaian begin mathsize 14px style x equals 1 half open parentheses 1 minus square root of 17 close parentheses end style dan begin mathsize 14px style x equals 1 half open parentheses 1 plus square root of 17 close parentheses end style, serta keduanya memenuhi syarat x not equal to negative 1.

Berikutnya, perhatikan pula perhitungan berikut!

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction end cell equals cell negative 3 semicolon space x not equal to negative 1 end cell row cell fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction times open parentheses x plus 1 close parentheses end cell equals cell negative 3 times open parentheses x plus 1 close parentheses end cell row cell x squared plus 2 x minus 1 end cell equals cell negative 3 x minus 3 end cell row cell x squared plus 5 x end cell equals cell negative 2 end cell row cell open parentheses x plus 5 over 2 close parentheses squared minus 25 over 4 end cell equals cell negative 2 end cell row cell open parentheses x plus 5 over 2 close parentheses squared end cell equals cell negative 2 plus 25 over 4 end cell row cell open parentheses x plus 5 over 2 close parentheses squared end cell equals cell 17 over 4 end cell row cell x plus 5 over 2 end cell equals cell plus-or-minus square root of 17 over 4 end root end cell row x equals cell negative 5 over 2 plus-or-minus 1 half square root of 17 end cell row x equals cell 1 half open parentheses negative 5 plus-or-minus square root of 17 close parentheses end cell end table

Dari persamaan begin mathsize 14px style fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction equals negative 3 end style, diperoleh penyelesaian begin mathsize 14px style x equals 1 half open parentheses negative 5 minus square root of 17 close parentheses end style dan begin mathsize 14px style x equals 1 half open parentheses negative 5 plus square root of 17 close parentheses end style, serta keduanya memenuhi syarat x not equal to negative 1.

Jika kita gambarkan dalam koordinat kartesius, antara fungsi begin mathsize 14px style f open parentheses x close parentheses equals open vertical bar fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction close vertical bar end style dan begin mathsize 14px style g open parentheses x close parentheses equals 3 end style akan berpotongan di empat titik seperti pada gambar berikut ini.

Dengan demikian, himpunan penyelesaian dari persamaan begin mathsize 14px style open vertical bar fraction numerator x squared plus 2 x minus 1 over denominator x plus 1 end fraction close vertical bar equals 3 end style adalah begin mathsize 12px style open curly brackets 1 half open parentheses negative 5 minus square root of 17 close parentheses comma blank 1 half open parentheses 1 minus square root of 17 close parentheses comma blank 1 half open parentheses 1 plus square root of 17 close parentheses comma blank 1 half open parentheses negative 5 plus square root of 17 close parentheses close curly brackets end style.

Jadi, jawaban yang tepat adalah C.

0

Roboguru

Himpunan penyelesaian bilangan real dari persamaan  adalah ...

Pembahasan Soal:

Dari definisi bentuk mutlak, kita punya

begin mathsize 14px style open vertical bar fraction numerator 2 x squared minus 4 x plus 4 over denominator 4 x minus 3 end fraction close vertical bar equals 2 open curly brackets table attributes columnalign right end attributes row cell fraction numerator 2 x squared minus 4 x plus 4 over denominator 4 x minus 3 end fraction equals 2 comma end cell cell jika blank fraction numerator 2 x squared minus 4 x plus 4 over denominator 4 x minus 3 end fraction greater or equal than 0 end cell row cell fraction numerator 2 x squared minus 4 x plus 4 over denominator 4 x minus 3 end fraction equals negative 2 comma end cell cell jika blank fraction numerator 2 x squared minus 4 x plus 4 over denominator 4 x minus 3 end fraction less than 0 end cell end table close end style  

Selanjutnya, kita cari masing-masing penyelesaian dari bentuk di atas.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 x squared minus 4 x plus 4 over denominator 4 x minus 3 end fraction end cell equals cell 2 semicolon x not equal to 2 end cell row cell fraction numerator 2 x squared minus 4 x plus 4 over denominator 4 x minus 3 end fraction times open parentheses 4 x minus 3 close parentheses end cell equals cell 2 times open parentheses 4 x minus 3 close parentheses end cell row cell 2 x squared minus 4 x plus 4 end cell equals cell 8 x minus 6 end cell row cell 2 x squared minus 12 x plus 10 end cell equals 0 row cell open parentheses 2 x squared minus 12 x plus 10 close parentheses divided by 2 end cell equals cell 0 divided by 2 end cell row cell x squared minus 6 x plus 5 end cell equals 0 row cell open parentheses x minus 1 close parentheses open parentheses x minus 5 close parentheses end cell equals 0 row x equals cell 1 blank atau blank x equals 5 end cell end table end style     

Dari persamaanbegin mathsize 14px style fraction numerator 2 x squared minus 4 x plus 4 over denominator 4 x minus 3 end fraction equals 2 end style kita dapatkan penyelesaiannya yakni x = 1 dan x = 5 serta keduanya memenuhi syarat bahwa begin mathsize 14px style x not equal to 3 over 4 end style .

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 x squared minus 4 x plus 4 over denominator 4 x minus 3 end fraction end cell equals cell negative 2 semicolon x not equal to 2 end cell row cell fraction numerator 2 x squared minus 4 x plus 4 over denominator 4 x minus 3 end fraction times open parentheses 4 x minus 3 close parentheses end cell equals cell negative 2 times open parentheses 4 x minus 3 close parentheses end cell row cell 2 x squared minus 4 x plus 4 end cell equals cell negative 8 x plus 6 end cell row cell 2 x squared plus 4 x end cell equals 2 row cell open parentheses 2 x squared plus 4 x close parentheses divided by 2 end cell equals cell 2 divided by 2 end cell row cell x squared plus 2 x end cell equals 1 row cell open parentheses x plus 1 close parentheses squared minus 1 end cell equals 1 row cell open parentheses x plus 1 close parentheses squared end cell equals 2 row cell x plus 1 end cell equals cell plus-or-minus square root of 2 end cell row x equals cell negative 1 plus-or-minus square root of 2 end cell end table end style      

Dari persamaan begin mathsize 14px style fraction numerator 2 x squared minus 4 x plus 4 over denominator 4 x minus 3 end fraction equals negative 2 end style kita dapatkan penyelesaiannya yakni begin mathsize 14px style x equals negative 1 minus square root of 2 blank end style dan begin mathsize 14px style x equals negative 1 plus square root of 2 end style serta keduanya memenuhi syarat bahwa undefined 

Jika kita gambarkan dalam koordinat kartesius, antara fungsi begin mathsize 14px style f open parentheses x close parentheses equals open vertical bar fraction numerator 2 x squared minus 4 x plus 4 over denominator 4 x minus 3 end fraction close vertical bar end style dan begin mathsize 14px style g open parentheses x close parentheses equals 2 end style bertemu di empat titik seperti gambar berikut ini.

Jadi, himpunan penyelesaian dari persamaan begin mathsize 14px style open vertical bar fraction numerator 2 x squared minus 4 x plus 4 over denominator 4 x minus 3 end fraction close vertical bar equals 2 end style adalah undefined sehingga jawaban yang tepat adalah B.

 

0

Roboguru

Hasil jumlah dari semua penyelesaian bilangan real dari persamaan  adalah ...

Pembahasan Soal:

Dari definisi bentuk mutlak, kita punya

begin mathsize 14px style open vertical bar fraction numerator x squared minus 3 x minus 4 over denominator x plus 2 end fraction close vertical bar equals 2 open curly brackets table attributes columnalign right end attributes row cell fraction numerator x squared minus 3 x minus 4 over denominator x plus 2 end fraction equals 2 comma end cell cell jika blank fraction numerator x squared minus 3 x minus 4 over denominator x plus 2 end fraction greater or equal than 0 end cell row cell fraction numerator x squared minus 3 x minus 4 over denominator x plus 2 end fraction equals negative 2 comma end cell cell jika blank fraction numerator x squared minus 3 x minus 4 over denominator x plus 2 end fraction less than 0 end cell end table close end style  

Selanjutnya, kita cari hasil jumlah penyelesaian bilangan real dari masing-masing bentuk di atas.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x squared minus 3 x minus 4 over denominator x plus 2 end fraction end cell equals cell 2 semicolon x not equal to negative 2 end cell row cell fraction numerator x squared minus 3 x minus 4 over denominator x plus 2 end fraction times open parentheses x plus 2 close parentheses end cell equals cell 2 times open parentheses x plus 2 close parentheses end cell row cell x squared minus 3 x minus 4 end cell equals cell 2 x plus 4 end cell row cell x squared minus 5 x minus 8 end cell equals 0 end table end style     

Dapat diperiksa bahwa persamaan begin mathsize 14px style x squared minus 5 x minus 8 equals 0 end style dengan a = 1,b = -5, dan c = -8 memiliki nilai diskriminan yang positif sehingga pasti memiliki dua penyelesaian bilangan real. Misalkan dua penyelesaian tersebut adalah begin mathsize 14px style straight x subscript 1 space dan space straight x subscript 2 end style maka

begin mathsize 14px style x subscript 1 plus x subscript 2 equals negative b over a equals negative fraction numerator negative 5 over denominator 1 end fraction equals 5 end style  

Kita lanjutkan ke bentuk yang kedua.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x squared minus 3 x minus 4 over denominator x plus 2 end fraction end cell equals cell negative 2 semicolon x not equal to negative 2 end cell row cell fraction numerator x squared minus 3 x minus 4 over denominator x plus 2 end fraction times open parentheses x plus 2 close parentheses end cell equals cell negative 2 times open parentheses x plus 2 close parentheses end cell row cell x squared minus 3 x minus 4 end cell equals cell negative 2 x minus 4 end cell row cell x squared minus x end cell equals 0 row cell x open parentheses x minus 1 close parentheses end cell equals 0 row x equals cell 0 blank atau blank x equals 1 end cell end table end style    

Dari bentuk kedua ini kita dapatkan hasil penjumlahan penyelesaiannya adalah 0 + 1 = 1.

Oleh karena itu, hasil jumlah semua penyelesaian bilangan real dari persamaan undefined adalah begin mathsize 14px style 5 plus 1 equals 6 end style 

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Himpunan penyelesaian bilangan real dari persamaan  adalah ...

Pembahasan Soal:

Dari definisi bentuk mutlak, kita punya

begin mathsize 14px style open vertical bar fraction numerator 2 x squared plus x minus 3 over denominator 2 x plus 4 end fraction close vertical bar equals 3 over 2 open curly brackets table attributes columnalign right end attributes row cell fraction numerator 2 x squared plus x minus 3 over denominator 2 x plus 4 end fraction equals 3 over 2 comma end cell cell jika blank fraction numerator 2 x squared plus x minus 3 over denominator 2 x plus 4 end fraction greater or equal than 0 end cell row cell fraction numerator 2 x squared plus x minus 3 over denominator 2 x plus 4 end fraction equals negative 3 over 2 comma end cell cell jika blank fraction numerator 2 x squared plus x minus 3 over denominator 2 x plus 4 end fraction less than 0 end cell end table close end style   

Selanjutnya, kita cari masing-masing penyelesaian dari bentuk di atas.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 x squared plus x minus 3 over denominator 2 x plus 4 end fraction end cell equals cell 3 over 2 semicolon x not equal to 2 end cell row cell fraction numerator 2 x squared plus x minus 3 over denominator 2 x plus 4 end fraction times 2 open parentheses 2 x plus 4 close parentheses end cell equals cell 3 over 2 times 2 open parentheses 2 x plus 4 close parentheses end cell row cell 4 x squared plus 2 x minus 6 end cell equals cell 6 x plus 12 end cell row cell 4 x squared minus 4 x end cell equals 18 row cell open parentheses 4 x squared minus 4 x close parentheses divided by 4 end cell equals cell 18 divided by 4 end cell row cell x squared minus x end cell equals cell 9 over 2 end cell row cell open parentheses x minus 1 half close parentheses squared minus 1 fourth end cell equals cell 9 over 2 end cell row cell open parentheses x minus 1 half close parentheses squared end cell equals cell 9 over 2 plus 1 fourth end cell row cell open parentheses x minus 1 half close parentheses squared end cell equals cell 19 over 4 end cell row cell x minus 1 half end cell equals cell plus-or-minus square root of 19 over 4 end root end cell row cell x minus 1 half end cell equals cell plus-or-minus 1 half square root of 19 end cell row x equals cell 1 half plus-or-minus 1 half square root of 19 end cell row x equals cell 1 half open parentheses 1 plus-or-minus square root of 19 close parentheses end cell row blank blank blank end table end style    

Dari persamaan begin mathsize 14px style fraction numerator 2 x squared plus x minus 3 over denominator 2 x plus 4 end fraction equals 3 over 2 end style, kita dapatkan penyelesaiannya yakni begin mathsize 14px style x equals 1 half open parentheses 1 minus square root of 19 close parentheses end style dan undefined serta keduanya memenuhi syarat bahwa x ≠ -2.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 x squared plus x minus 3 over denominator 2 x plus 4 end fraction end cell equals cell negative 3 over 2 semicolon x not equal to 2 end cell row cell fraction numerator 2 x squared plus x minus 3 over denominator 2 x plus 4 end fraction times 2 open parentheses 2 x plus 4 close parentheses end cell equals cell negative 3 over 2 times 2 open parentheses 2 x plus 4 close parentheses end cell row cell 4 x squared plus 2 x minus 6 end cell equals cell negative 6 x minus 12 end cell row cell 4 x squared plus 8 x plus 6 end cell equals 0 end table end style    

Perhatikan bahwa persamaan begin mathsize 14px style 4 x squared plus 8 x plus 6 equals 0 end style dengan begin mathsize 14px style a equals 4 comma b equals 8 comma end style dan begin mathsize 14px style c equals 6 end style memiliki nilai diskriminan yang negatif yakni

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row D equals cell b squared minus 4 a c end cell row blank equals cell 8 squared minus 4 times 4 times 6 end cell row blank equals cell 64 minus 96 end cell row blank equals cell negative 32 end cell end table end style   

Artinya, dari persamaan begin mathsize 14px style fraction numerator 2 x squared plus x minus 3 over denominator 2 x plus 4 end fraction equals negative 3 over 2 end style, kita tidak mendapatkan penyelesaian bilangan real.

Jika kita gambarkan dalam koordinat kartesius, antara fungsi begin mathsize 14px style f open parentheses x close parentheses equals open vertical bar fraction numerator 2 x squared plus x minus 3 over denominator 2 x plus 4 end fraction close vertical bar end style dan begin mathsize 14px style g open parentheses x close parentheses equals 3 over 2 end style bertemu di dua titik seperti gambar berikut ini.

Jadi, himpunan penyelesaian dari persamaan begin mathsize 14px style open vertical bar fraction numerator 2 x squared plus x minus 3 over denominator 2 x plus 4 end fraction close vertical bar equals 3 over 2 end style adalah undefined sehingga jawaban yang tepat adalah D.

0

Roboguru

Diketahui salah satu penyelesaian dari persamaan  adalah 2. Jika a negatif, maka himpunan penyelesaian bilangan real persamaan tersebut adalah ....

Pembahasan Soal:

Diketahui x = 2 adalah salah satu persamaan begin mathsize 14px style open vertical bar fraction numerator x minus a over denominator x plus 2 end fraction close vertical bar equals 2 end style maka kita dapatkan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator x minus a over denominator x plus 1 end fraction close vertical bar end cell equals 2 row cell open vertical bar fraction numerator 2 minus a over denominator 2 plus 1 end fraction close vertical bar end cell equals 2 row cell open vertical bar fraction numerator 2 minus a over denominator 3 end fraction close vertical bar end cell equals 2 row cell 1 third open vertical bar 2 minus a close vertical bar end cell equals 2 row cell 1 third open vertical bar 2 minus a close vertical bar times 3 end cell equals cell 2 times 3 end cell row cell open vertical bar 2 minus a close vertical bar end cell equals 6 row blank blank cell table attributes columnalign right end attributes row cell 2 minus a equals 6 end cell atau cell 2 minus a equals negative 6 end cell row cell negative a equals 4 end cell blank cell negative a equals negative 8 end cell row cell a equals negative 4 end cell blank cell a equals 8 end cell end table end cell end table end style     

Karena a negatif, maka haruslah a = -4 sehingga persamaan tersebut dapat kita tuliskan kembali sebagai berikut.

begin mathsize 14px style open vertical bar fraction numerator x plus 4 over denominator x plus 1 end fraction close vertical bar equals 2 open curly brackets table attributes columnalign right end attributes row cell fraction numerator x plus 4 over denominator x plus 1 end fraction equals 2 comma end cell cell jika blank fraction numerator x plus 4 over denominator x plus 1 end fraction greater or equal than 0 end cell row cell fraction numerator x plus 4 over denominator x plus 1 end fraction equals negative 2 comma end cell cell jika blank fraction numerator x plus 4 over denominator x plus 1 end fraction less than 0 end cell end table close end style   

Selanjutnya, kita cari masing-masing penyelesaian dari bentuk di atas.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x plus 4 over denominator x plus 1 end fraction end cell equals cell 2 semicolon x not equal to negative 1 end cell row cell fraction numerator x plus 4 over denominator x plus 1 end fraction times open parentheses x plus 1 close parentheses end cell equals cell 2 times open parentheses x plus 1 close parentheses end cell row cell x plus 4 end cell equals cell 2 x plus 2 end cell row cell negative x end cell equals cell negative 2 end cell row x equals 2 row blank blank blank row cell fraction numerator x plus 4 over denominator x plus 1 end fraction end cell equals cell negative 2 semicolon x not equal to negative 1 end cell row cell fraction numerator x plus 4 over denominator x plus 1 end fraction times open parentheses x plus 1 close parentheses end cell equals cell negative 2 times open parentheses x plus 1 close parentheses end cell row cell x plus 4 end cell equals cell negative 2 x minus 2 end cell row cell 3 x end cell equals cell negative 6 end cell row x equals cell negative 2 end cell end table end style    

Kita dapatkan bahwa himpunan penyelesaian dari persamaan begin mathsize 14px style open vertical bar fraction numerator x plus 4 over denominator x plus 1 end fraction close vertical bar equals 2 end style adalah {-2,2}.
Jadi, jawaban yang tepat adalah D. 

0

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