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

Pertanyaan

Himpunan bilangan real x yang merupakan 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 comma space x not equal to negative 1 end style adalah ....

  1. begin mathsize 12px style open curly brackets 1 half open parentheses negative 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

  2. begin mathsize 12px style open curly brackets 1 half open parentheses 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 5 plus square root of 17 close parentheses close curly brackets end style

  3. 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

  4. 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 negative 5 plus square root of 17 close parentheses close curly brackets end style

  5. begin mathsize 12px style open curly brackets 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 close curly brackets end style

D. Natalia

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

Pembahasan

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.

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