Roboguru

Tentukan himpunan penyelesaian dari pertidaksamaan-pertidaksamaan: 1.

Pertanyaan

Tentukan himpunan penyelesaian dari pertidaksamaan-pertidaksamaan:

1. open vertical bar 2 x minus 5 close vertical bar less or equal than 5

Pembahasan Soal:

Ingat sifat pertidaksamaan nilai mutlak:

open vertical bar f left parenthesis x right parenthesis close vertical bar less or equal than c left right arrow negative c less or equal than f left parenthesis x right parenthesis less or equal than c

Sehingga, diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell negative 5 end cell less or equal than cell 2 x minus 5 less or equal than 5 end cell row cell negative 5 plus 5 end cell less or equal than cell 2 x less or equal than 5 plus 5 end cell row 0 less or equal than cell 2 x less or equal than 10 end cell row 0 less or equal than cell x less or equal than 5 end cell end table

Jadi, himpunan penyelesaiannya adalah open curly brackets x vertical line table attributes columnalign right center left columnspacing 0px end attributes row blank blank 0 end table table attributes columnalign right center left columnspacing 0px end attributes row blank less or equal than blank end table 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 less or equal than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank comma end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table 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 element of blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank straight real numbers end table close curly brackets

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Bella

Mahasiswa/Alumni UIN Syarif Hidayatullah Jakarta

Terakhir diupdate 13 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai x yang memenuhi pertidaksamaan  adalah...

Pembahasan Soal:

Diketahui pertidaksamaan vertical line x minus 2 vertical line less than 3

Maka

table attributes columnalign right center left columnspacing 0px end attributes row cell negative 3 end cell less than cell x minus 2 less than 3 end cell row cell negative 3 plus 2 end cell less than cell x minus 2 plus 2 less than 3 plus 2 end cell row cell negative 1 end cell less than cell x less than 5 end cell end table 

Jadi, jawaban yang benar adalah C.

0

Roboguru

Tentukan himpunan penyelesaian dari pertidaksamaan nilai mutlak berikut: 2.

Pembahasan Soal:

Ingat kembali:

f open parentheses x close parentheses less or equal than c space rightwards arrow space minus c less or equal than f open parentheses x close parentheses less or equal than c  

Diketahui pertidaksamaan open vertical bar x plus 3 close vertical bar less or equal than 1, maka:

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

Jadi, himpunan penyelesaian dari open vertical bar x plus 3 close vertical bar less or equal than 1 adalah negative 4 less or equal than x less or equal than negative 2.

1

Roboguru

Tentukan himpunan penyelesaian pertidaksamaan berikut

Pembahasan Soal:

Ingat kembali nilai mutlak

open vertical bar x close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell x comma space x greater or equal than 0 end cell row cell negative x comma space x less than 0 end cell end table close 

Diperhatikan kembali

open vertical bar x minus 3 close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell x minus 3 comma space x minus 3 greater or equal than 0 rightwards double arrow x greater or equal than 3 end cell row cell negative left parenthesis x minus 3 right parenthesis comma space x minus 3 less than 0 rightwards double arrow x less than 3 end cell end table close 

Diperoleh perhitungan untuk dua kasus

  • untuk x greater or equal than 3 

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 3 end cell less or equal than 4 row x less or equal than 7 end table 

  • untuk x less than 3 

table attributes columnalign right center left columnspacing 0px end attributes row cell negative left parenthesis x minus 3 right parenthesis end cell less or equal than 4 row cell negative x plus 3 end cell less or equal than 4 row cell negative x end cell less or equal than 1 row x greater or equal than cell negative 1 end cell end table 

Dengan demikian, diperoleh hasilnya yaitu x less or equal than 7 atau x greater or equal than negative 1.

0

Roboguru

Tentukan himpunan penyelesaian dari persamaan dan pertidaksaman nilai mutlak berikut. c.

Pembahasan Soal:

Ingat 

open vertical bar f open parentheses x close parentheses close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell space space space f open parentheses x close parentheses comma space space space j i k a space space f open parentheses x close parentheses greater or equal than 0 end cell row cell negative f open parentheses x close parentheses comma space space space j i k a space space f open parentheses x close parentheses less than 0 end cell end table close 

Sehingga, pada soal kita

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x over 3 minus 2 close vertical bar space end cell rightwards arrow cell open parentheses x over 3 minus 2 close parentheses space j i k a space x greater or equal than 6 end cell row cell space space space space space space space space space space space end cell rightwards arrow cell open parentheses 2 minus x over 3 close parentheses space j i k a space x less than 6 end cell row blank blank blank row cell Kondisi space 1 space colon space x end cell greater or equal than 6 row cell x over 3 minus 2 end cell less or equal than 10 row cell x over 3 minus 12 end cell less or equal than 0 row cell fraction numerator x minus 36 over denominator 3 end fraction end cell less or equal than 0 row x less or equal than 36 row blank blank cell Diiriskan space dengan space syarat space kondisi space end cell row x greater or equal than cell 6 space end cell row blank blank cell Sehingga space irisannya space menjadi space end cell row 6 less or equal than cell x less or equal than 36 end cell row blank blank blank row cell Kondisi space 2 space colon space x end cell less than 6 row cell 2 minus x over 3 end cell less or equal than 10 row cell fraction numerator negative x over denominator 3 end fraction end cell less or equal than 8 row cell negative x end cell less or equal than 24 row x greater or equal than cell negative 24 end cell row blank blank cell Diiriskan space dengan space syarat space kondisi space 2 colon end cell row cell negative 24 end cell less or equal than cell x less than 6 end cell row blank blank blank row blank blank blank row blank blank space end table end style

 

Jadi daerah gabungan penyelesaian adalah :

begin mathsize 14px style negative 24 less or equal than x less or equal than 36 end style  

0

Roboguru

Selesaikan setiap PtLSVNM berikut.

Pembahasan Soal:

Bentuk open vertical bar 3 x plus 4 close vertical bar less than 8 mempunyai penyelesaian awal:


table attributes columnalign right center left columnspacing 0px end attributes row cell negative 8 space end cell less than cell 3 x plus 4 less than 8 end cell row blank left right arrow cell negative 8 minus 4 less than space space space 3 x space space space less than space 8 minus 4 end cell row blank left right arrow cell space space space space minus 12 less than space space space 3 x space space space less than space 4 end cell row blank left right arrow cell space space minus 12 over 3 less than space space fraction numerator 3 x over denominator 3 end fraction space space space less than space 4 over 3 end cell row blank left right arrow cell space space space space space minus 4 space less than space space space space x space space space space space less than space 4 over 3 end cell end table


Jadi penyelesaian akhir: negative 4 less than x less than 4 over 3.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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