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Penyelesaian pertidaksamaan  adalah ....

Pertanyaan

Penyelesaian pertidaksamaan square root of 2 x minus 1 end root greater or equal than x minus 2 adalah ....

  1. 1 less or equal than straight x less or equal than 5 

  2. 1 half less or equal than straight x less or equal than 5 

  3. negative 1 half less or equal than straight x less or equal than 5 

  4. straight x less than 1 space atau space straight x greater than 5 

  5. straight x less or equal than negative 1 half space atau space straight x greater or equal than 5 

Pembahasan Soal:

Pertidaksamaan bentuk akar.

i) kuadratkan kedua ruas

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of 2 straight x minus 1 end root end cell greater or equal than cell straight x minus 2 end cell row cell left parenthesis square root of 2 straight x minus 1 end root right parenthesis squared end cell greater or equal than cell left parenthesis straight x minus 2 right parenthesis squared space end cell row cell 2 straight x minus 1 end cell greater or equal than cell straight x squared minus 4 straight x plus 4 end cell row 0 greater or equal than cell straight x squared minus 6 straight x plus 5 end cell row cell straight x squared minus 6 straight x plus 5 end cell less or equal than 0 row cell left parenthesis straight x minus 5 right parenthesis left parenthesis straight x minus 1 right parenthesis end cell less or equal than 0 end table  

Uji coba garis bilangan dengan batas straight x equals 1 space dan space straight x equals 5 sehingga diperoleh :

table row cell plus plus plus end cell blank cell negative negative negative end cell blank cell plus plus plus end cell row blank 1 blank 5 blank end table 

Karena table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell left parenthesis straight x minus 5 right parenthesis left parenthesis straight x minus 1 right parenthesis 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 blank blank 0 end table maka yang memenuhi adalah 1 less or equal than x less or equal than 5. 

ii) solusi syarat bentuk akar yaitu ;

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 straight x minus 1 end cell greater or equal than 0 row cell 2 straight x end cell greater or equal than 1 row straight x greater or equal than cell 1 half end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell HP 1 end cell identical to cell 1 less or equal than straight x less or equal than 5 end cell row cell HP 1 end cell identical to cell straight x greater or equal than 1 half end cell row cell HP 1 intersection HP 1 end cell identical to cell 1 half less or equal than straight x less or equal than 5 end cell end table 

Jadi, Penyelesaian pertidaksamaan square root of 2 x minus 1 end root greater or equal than x minus 2 adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half 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 blank blank straight 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. end table 

Oleh karena itu, Jawaban yang benar adalah B

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 05 Mei 2021

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