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Tentukan himpunan penyelesaian dari: c. (x2−3x+2)3​≤(x2−4x+3)5​

Pertanyaan

Tentukan himpunan penyelesaian dari:

c. fraction numerator 3 over denominator open parentheses x squared minus 3 x plus 2 close parentheses end fraction less or equal than fraction numerator 5 over denominator open parentheses x squared minus 4 x plus 3 close parentheses end fraction 

S. Dwi

Master Teacher

Jawaban terverifikasi

Jawaban

himpunan penyelesaiannya adalah open curly brackets right enclose x 1 less than x less than 3 close curly brackets 

Pembahasan

Himpunan penyelesaian pertidaksamaan dapat diselesaikan dengan langkah:

Langkah 1. Menentukan pembuat nol

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 3 over denominator straight x squared minus 3 straight x plus 2 end fraction end cell less or equal than cell fraction numerator 5 over denominator straight x squared minus 4 straight x plus 3 end fraction end cell row cell fraction numerator 3 over denominator straight x squared minus 3 straight x plus 2 end fraction minus fraction numerator 5 over denominator straight x squared minus 4 straight x plus 3 end fraction end cell less or equal than 0 row cell fraction numerator 3 open parentheses straight x squared minus 4 straight x plus 3 close parentheses minus 5 open parentheses straight x squared minus 3 straight x plus 2 close parentheses over denominator open parentheses straight x squared minus 3 straight x plus 2 close parentheses open parentheses straight x squared minus 4 straight x plus 3 close parentheses end fraction end cell less or equal than 0 row cell fraction numerator 3 straight x squared minus 12 straight x plus 9 minus 5 straight x squared plus 15 straight x minus 10 over denominator open parentheses straight x squared minus 3 straight x plus 2 close parentheses open parentheses straight x squared minus 4 straight x plus 3 close parentheses end fraction end cell less or equal than 0 row cell fraction numerator negative 2 straight x squared plus 3 straight x minus 1 over denominator open parentheses straight x squared minus 3 straight x plus 2 close parentheses open parentheses straight x squared minus 4 straight x plus 3 close parentheses end fraction end cell less or equal than 0 row cell fraction numerator open parentheses negative 2 straight x plus 1 close parentheses open parentheses straight x minus 1 close parentheses over denominator open parentheses straight x minus 1 close parentheses open parentheses straight x minus 2 close parentheses open parentheses straight x minus 1 close parentheses open parentheses straight x minus 3 close parentheses end fraction end cell less or equal than cell 0 space bold difaktorkan end cell end table  

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open table attributes columnalign right end attributes row cell straight x subscript 1 equals 1 half space end cell row cell straight x subscript 2 equals 1 end cell row cell straight x subscript 3 equals 2 space end cell row cell straight x subscript 4 equals 3 space end cell end table close curly brackets bold space bold pembuat bold space bold nol bold space bold left parenthesis bold 0 bold right parenthesis end cell end table  

Langkah 2. Menguji titik disekitaran pembuat nol

Pembuat nol pada garis bilangan dibagi menjadi 5 bagian. Selanjutnya, pilih sembarang titik/nilai pada masing-masing bagian

begin mathsize 12px style straight x equals 0 rightwards double arrow fraction numerator open parentheses negative 2 times 0 plus 1 close parentheses open parentheses 0 minus 1 close parentheses over denominator open parentheses 0 minus 1 close parentheses open parentheses 0 minus 2 close parentheses open parentheses 0 minus 1 close parentheses open parentheses 0 minus 3 close parentheses end fraction equals fraction numerator negative 1 over denominator 6 end fraction less or equal than 0 straight x equals 1 comma 5 rightwards double arrow fraction numerator open parentheses negative 2 times 1 comma 5 plus 1 close parentheses open parentheses 1 comma 5 minus 1 close parentheses over denominator open parentheses 1 comma 5 minus 1 close parentheses open parentheses 1 comma 5 minus 2 close parentheses open parentheses 1 comma 5 minus 1 close parentheses open parentheses 1 comma 5 minus 3 close parentheses end fraction equals fraction numerator 1 comma 5 over denominator 0 comma 18 end fraction less or equal than 0 straight x equals 4 rightwards double arrow fraction numerator open parentheses negative 2 times 4 plus 1 close parentheses open parentheses 4 minus 1 close parentheses over denominator open parentheses 4 minus 1 close parentheses open parentheses 4 minus 2 close parentheses open parentheses 4 minus 1 close parentheses open parentheses 4 minus 3 close parentheses end fraction equals fraction numerator negative 21 over denominator 18 end fraction less or equal than 0 end style

Langkah 3. Menentukan daerah himpunan penyelesaian

Setelah diuji maka didapat daerah penyelesaian dari sistem pertidaksamaan rasional. Dengan demikian, himpunan penyelesaiannya adalah open curly brackets right enclose x 1 less than x less than 3 close curly brackets 

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