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Himpunan penyelesaian pertidaksamaan ∣∣​x−4x+3​∣∣​<2 adalah ...

Pertanyaan

Himpunan penyelesaian pertidaksamaan open vertical bar fraction numerator x plus 3 over denominator x minus 4 end fraction close vertical bar less than 2 adalah ...

  1. open curly brackets right enclose x space end enclose space x less than 5 over 3 close curly brackets

  2. open curly brackets right enclose x space end enclose space x greater than 11 close curly brackets

  3. open curly brackets right enclose x space end enclose space 5 over 3 less than x less than 11 close curly brackets

  4. open curly brackets right enclose x space end enclose space x less than 5 over 3 space atau space x greater than 11 close curly brackets

  5. open curly brackets right enclose x space end enclose space x greater than 5 over 3 close curly brackets intersection open curly brackets right enclose x space end enclose space x less than 11 close curly brackets

Pembahasan Soal:

Sifat pertidaksamaan nilai mutlak yaitu:

table attributes columnalign center center left center center end attributes row cell fraction numerator open vertical bar f open parentheses x close parentheses close vertical bar over denominator open vertical bar g open parentheses x close parentheses close vertical bar end fraction less than a end cell left right double arrow cell open vertical bar f open parentheses x close parentheses close vertical bar less than a open vertical bar g open parentheses x close parentheses close vertical bar end cell space space row space left right double arrow cell open vertical bar f open parentheses x close parentheses close vertical bar less than open vertical bar a times g open parentheses x close parentheses close vertical bar end cell space space row space left right double arrow cell open square brackets f open parentheses x close parentheses plus a times g open parentheses x close parentheses close square brackets open square brackets f open parentheses x close parentheses minus a times g open parentheses x close parentheses close square brackets end cell less than 0 end table

Sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator x plus 3 over denominator x minus 4 end fraction close vertical bar end cell less than 2 row cell open vertical bar x plus 3 close vertical bar end cell less than cell 2 open vertical bar x minus 4 close vertical bar end cell row cell open vertical bar x plus 3 close vertical bar end cell less than cell open vertical bar 2 open parentheses x minus 4 close parentheses close vertical bar end cell row cell open vertical bar x plus 3 close vertical bar end cell less than cell open vertical bar 2 x minus 8 close vertical bar end cell row cell open square brackets open parentheses x plus 3 close parentheses plus open parentheses 2 x minus 8 close parentheses close square brackets open square brackets open parentheses x plus 3 close parentheses minus open parentheses 2 x minus 8 close parentheses close square brackets end cell less than 0 row cell open parentheses 3 x minus 5 close parentheses open parentheses x minus 2 x plus 3 plus 8 close parentheses end cell less than 0 row cell open parentheses 3 x minus 5 close parentheses open parentheses negative x plus 11 close parentheses end cell less than 0 end table

Pembuat nol:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x minus 5 end cell equals 0 row cell 3 x end cell equals 5 row x equals cell 5 over 3 end cell end table

atau

table attributes columnalign right center left columnspacing 0px end attributes row cell negative x plus 11 end cell equals 0 row cell negative x end cell equals cell negative 11 end cell row x equals 11 end table

Uji titik:

Untuk interval x greater than 11, ambil x equals 12, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 3 x minus 5 close parentheses open parentheses negative x plus 11 close parentheses end cell equals cell open parentheses 3 open parentheses 12 close parentheses minus 5 close parentheses open parentheses negative 12 plus 11 close parentheses end cell row blank equals cell open parentheses 36 minus 5 close parentheses open parentheses negative 1 close parentheses end cell row blank equals cell 31 open parentheses negative 1 close parentheses end cell row blank equals cell negative 31 end cell end table

Untuk interval 5 over 3 less than x less than 11, ambil x equals 5, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 3 x minus 5 close parentheses open parentheses negative x plus 11 close parentheses end cell equals cell open parentheses 3 open parentheses 5 close parentheses minus 5 close parentheses open parentheses negative 5 plus 11 close parentheses end cell row blank equals cell open parentheses 15 minus 5 close parentheses open parentheses 6 close parentheses end cell row blank equals cell 5 open parentheses 6 close parentheses end cell row blank equals 30 end table

Untuk interval x less than 5 over 3, ambil x equals 0, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 3 x minus 5 close parentheses open parentheses negative x plus 11 close parentheses end cell equals cell open parentheses 3 open parentheses 0 close parentheses minus 5 close parentheses open parentheses negative open parentheses 0 close parentheses plus 11 close parentheses end cell row blank equals cell open parentheses 0 minus 5 close parentheses open parentheses 11 close parentheses end cell row blank equals cell open parentheses negative 5 close parentheses open parentheses 11 close parentheses end cell row blank equals cell negative 55 end cell end table

Diperoleh garis bilangannya yaitu:

Karena pertidaksamaan bertanda "less than", maka daerah penyelesaiannya berada pada interval yang bertanda open parentheses minus close parentheses.

Himpunan penyelesaian pertidaksamaan open vertical bar fraction numerator x plus 3 over denominator x minus 4 end fraction close vertical bar less than 2 adalah open curly brackets right enclose x space end enclose space x less than 5 over 3 space atau space x greater than 11 close curly brackets.

Jadi, jawaban yang tepat adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Faiz

Terakhir diupdate 06 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Selesaikan dan tulislah himpunan penyelesaiannya. ∣∣​3+x6−5x​∣∣​≤21​

Pembahasan Soal:

Pertidaksamaan open vertical bar fraction numerator 6 minus 5 x over denominator 3 plus x end fraction close vertical bar less or equal than 1 half memenuhi bentuk open vertical bar a over b close vertical bar less or equal than c dengan c equals 1 half greater than 0, maka bentuk tersebut dapat diubah menjadi:


table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets 2 left parenthesis 6 minus 5 x right parenthesis minus left parenthesis x plus 3 right parenthesis close square brackets open square brackets 2 left parenthesis 6 minus 5 x right parenthesis plus left parenthesis x plus 3 right parenthesis close square brackets end cell less or equal than 0 row cell open parentheses negative 11 x plus 9 close parentheses open parentheses negative 9 x plus 15 close parentheses end cell less or equal than 0 row cell open parentheses 11 x minus 9 close parentheses open parentheses 9 x minus 15 close parentheses end cell less or equal than 0 row cell 3 open parentheses 11 x minus 9 close parentheses open parentheses 3 x minus 5 close parentheses end cell less or equal than 0 end table


Jadi penyelesaiannya adalah 9 over 11 less or equal than x less or equal than 5 over 3.

0

Roboguru

Tentukan himpunan penyelesaian dari setiap PtNM berikut. c. ∣x−1∣∣x−2∣+3​−1<0

Pembahasan Soal:

Syarat:

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 1 end cell not equal to 0 row x not equal to cell 1 space..... space left parenthesis 1 right parenthesis end cell end table

Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator open vertical bar x minus 2 close vertical bar plus 3 over denominator open vertical bar x minus 1 close vertical bar end fraction minus 1 end cell less than 0 row cell fraction numerator open vertical bar x minus 2 close vertical bar plus 3 minus open vertical bar x minus 1 close vertical bar over denominator open vertical bar x minus 1 close vertical bar end fraction end cell less than 0 row cell open vertical bar x minus 2 close vertical bar plus 3 minus open vertical bar x minus 1 close vertical bar end cell less than 0 row cell open vertical bar x minus 2 close vertical bar end cell less than cell open vertical bar x minus 1 close vertical bar minus 3 end cell row cell open vertical bar x minus 2 close vertical bar squared end cell less than cell open parentheses open vertical bar x minus 1 close vertical bar minus 3 close parentheses squared end cell end table

Error converting from MathML to accessible text.
table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 4 x plus 4 minus x squared plus 2 x minus 1 minus 9 end cell less than cell negative 6 open vertical bar x minus 1 close vertical bar end cell row cell negative 2 x minus 6 end cell less than cell negative 6 open vertical bar x minus 1 close vertical bar end cell row cell 6 open vertical bar x minus 1 close vertical bar end cell less than cell 2 x minus 6 end cell row cell open parentheses 6 open vertical bar x minus 1 close vertical bar close parentheses squared end cell less than cell open parentheses 2 x minus 6 close parentheses squared end cell row cell 36 left parenthesis x squared minus 2 x plus 1 right parenthesis end cell less than cell 4 x squared minus 24 x plus 36 end cell row cell 36 x squared minus 72 x plus 36 minus 4 x squared plus 24 x minus 36 end cell less than 0 row cell 32 x squared minus 48 x end cell less than 0 row cell 2 x squared minus 3 x end cell less than 0 row cell x left parenthesis 2 x minus 3 right parenthesis end cell less than 0 row 0 less than cell x less than 3 over 2 space... space left parenthesis 2 right parenthesis end cell end table

Iriskan (1) dan (2) sehingga penyelesaiannya menjadi 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 than blank end table table attributes columnalign right center left columnspacing 0px end attributes row x less than cell 3 over 2 end cell end table comma x not equal to 1.

Jadi, himpunan penyelesaian pertidaksamaan fraction numerator open vertical bar x minus 2 close vertical bar plus 3 over denominator open vertical bar x minus 1 close vertical bar end fraction minus 1 less than 0 adalah open curly brackets table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell right enclose x end cell end table 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 than blank end table table attributes columnalign right center left columnspacing 0px end attributes row x less than cell 3 over 2 end cell end table comma space x not equal to 1 close curly brackets 

0

Roboguru

Selesaikan dan tuliskan himpunan penyelesaiannya dari PtNM berikut. c. ∣∣​2x−3x+2​∣∣​<4

Pembahasan Soal:

Syarat:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 3 end cell not equal to 0 row cell 2 x end cell not equal to 3 row x not equal to cell 3 over 2 space... space left parenthesis 1 right parenthesis end cell end table

Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator x plus 2 over denominator 2 x minus 3 end fraction close vertical bar end cell less than 4 row cell fraction numerator open vertical bar x plus 2 close vertical bar over denominator open vertical bar 2 x minus 3 close vertical bar end fraction end cell less than 4 row cell open vertical bar x plus 2 close vertical bar end cell less than cell 4 open vertical bar 2 x minus 3 close vertical bar end cell row cell open vertical bar x plus 2 close vertical bar end cell less than cell open vertical bar 8 x minus 12 close vertical bar end cell row cell left parenthesis left parenthesis x plus 2 right parenthesis plus left parenthesis 8 x minus 12 right parenthesis right parenthesis left parenthesis left parenthesis x plus 2 right parenthesis minus left parenthesis 8 x minus 12 right parenthesis right parenthesis end cell less than 0 row cell left parenthesis x plus 2 plus 8 x minus 12 right parenthesis left parenthesis x plus 2 minus 8 x plus 12 right parenthesis end cell less than 0 row cell left parenthesis 9 x minus 10 right parenthesis left parenthesis negative 7 x plus 14 right parenthesis end cell less than 0 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 than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 10 over 9 end cell 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 atau 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 greater than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 14 over 7 end cell end table x table attributes columnalign right center left columnspacing 0px end attributes row blank less than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 10 over 9 end cell 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 atau 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 greater than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table space... space left parenthesis 2 right parenthesis

Iriskan (1) dan (2) sehingga penyelesaiannya menjadi x table attributes columnalign right center left columnspacing 0px end attributes row blank less than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 10 over 9 end cell 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 atau 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 greater than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table.

Jadi, himpunan penyelesaian dari open vertical bar fraction numerator x plus 2 over denominator 2 x minus 3 end fraction close vertical bar less than 4 adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open curly brackets right enclose x x less than 10 over 9 space atau space x greater than 2 close curly brackets end cell end table 

0

Roboguru

Selesaikan dan tuliskan himpunan penyelesaiannya dari PtNM berikut. b. ∣∣​3+x6−5x​∣∣​​≤​21​​

Pembahasan Soal:

Syarat:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 plus x end cell not equal to 0 row x not equal to cell negative 3 space... space left parenthesis 1 right parenthesis end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator 6 minus 5 x over denominator 3 plus x end fraction close vertical bar end cell less or equal than cell 1 half end cell row cell fraction numerator open vertical bar 6 minus 5 x close vertical bar over denominator open vertical bar 3 plus x close vertical bar end fraction end cell less or equal than cell 1 half end cell row cell 2 cross times open vertical bar 6 minus 5 x close vertical bar end cell less or equal than cell open vertical bar 3 plus x close vertical bar end cell row cell open vertical bar 12 minus 10 x close vertical bar end cell less or equal than cell open vertical bar 3 plus x close vertical bar end cell row cell left parenthesis left parenthesis 12 minus 10 x right parenthesis plus left parenthesis 3 plus x right parenthesis right parenthesis left parenthesis left parenthesis 12 minus 10 x minus left parenthesis 3 plus x right parenthesis right parenthesis end cell less or equal than 0 row cell left parenthesis 12 minus 10 x plus 3 plus x right parenthesis left parenthesis 12 minus 10 x minus 3 minus x right parenthesis end cell less or equal than 0 row cell left parenthesis negative 9 x plus 15 right parenthesis left parenthesis negative 11 x plus 9 right parenthesis end cell less or equal than 0 row cell 9 over 11 end cell less or equal than cell x less or equal than 15 over 9 end cell row cell 9 over 11 end cell less or equal than cell x less or equal than 5 over 3 space... space left parenthesis 2 right parenthesis end cell end table

Iriskan (1) dan (2) sehingga penyelesaian menjadi table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 9 over 11 end cell 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 x less or equal than cell 5 over 3 end cell end table.

Jadi, himpunan penyelesaian pertidaksamaan table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator 6 minus 5 x over denominator 3 plus x end fraction close vertical bar end cell less or equal than cell 1 half end cell end table adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open curly brackets right enclose x 9 over 11 less or equal than x less or equal than 5 over 3 close curly brackets end cell end table 

0

Roboguru

Selesaikan dan tuliskan himpunan penyelesaian. ∣∣​2x−3x+2​∣∣​<4

Pembahasan Soal:

Bentuk open vertical bar fraction numerator x plus 2 over denominator 2 x minus 3 end fraction close vertical bar less than 4 memenuhi open vertical bar a over b close vertical bar less than c dengan c equals 4 greater than 0, maka:


table attributes columnalign right center left columnspacing 0px end attributes row cell open square brackets left parenthesis x plus 2 right parenthesis plus 4 left parenthesis 2 x minus 3 right parenthesis close square brackets open square brackets left parenthesis x plus 2 right parenthesis minus 4 left parenthesis 2 x minus 3 right parenthesis close square brackets end cell less than 0 row cell open square brackets left parenthesis x plus 2 right parenthesis plus left parenthesis 8 x minus 12 right parenthesis close square brackets open square brackets left parenthesis x plus 2 right parenthesis minus left parenthesis 8 x minus 12 right parenthesis close square brackets end cell less than 0 row cell left parenthesis 9 x minus 10 right parenthesis left parenthesis negative 7 x plus 14 right parenthesis end cell less than 0 row cell left parenthesis 9 x minus 10 right parenthesis left parenthesis 7 x minus 14 right parenthesis end cell greater than 0 row x less than cell 10 over 9 space atau space x greater than 2 end cell end table


Jadi, HP equals open curly brackets right enclose x space x less than 10 over 9 space atau space x greater than 2 close curly brackets.

1

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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