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Jika x 1 ​ dan x 2 ​ adalah akar-akar persamaan kuadrat x 2 − 8 x + 9 = 0. Jumlah tak hingga deret

Jika dan adalah akar-akar persamaan kuadrat Jumlah tak hingga deret open parentheses 1 over straight x subscript 1 close parentheses plus open parentheses 1 over straight x subscript 2 close parentheses plus open parentheses 1 over straight x subscript 1 close parentheses squared plus open parentheses 1 over straight x subscript 2 close parentheses squared plus open parentheses 1 over straight x subscript 1 close parentheses cubed plus open parentheses 1 over straight x subscript 2 close parentheses cubed plus horizontal ellipsis equals horizontal ellipsis

  1. 5

  2. 4

  3. 3

  4. 2

  5. 1

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N. Ayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Padang

Jawaban terverifikasi

Pembahasan

Karena dan adalah akar-akar persamaan kuadrat maka Jumlah tak hingga deretnya adalah... Jadi jumlah tak hingga di atas adalah

Karena straight x subscript 1 dan straight x subscript 2 adalah akar-akar persamaan kuadrat straight x squared minus 8 straight x plus 9 equals 0 maka

straight x subscript 1 plus straight x subscript 2 equals 8  straight x subscript 1 straight x subscript 2 equals 9

 

Jumlah tak hingga deretnya adalah...

open parentheses 1 over straight x subscript 1 close parentheses plus open parentheses 1 over straight x subscript 1 close parentheses squared plus open parentheses 1 over straight x subscript 1 close parentheses cubed plus horizontal ellipsis plus open parentheses 1 over straight x subscript 2 close parentheses plus open parentheses 1 over straight x subscript 2 close parentheses squared plus open parentheses 1 over straight x subscript 2 close parentheses cubed plus horizontal ellipsis equals...  stack stack open parentheses 1 over straight x subscript 1 close parentheses plus open parentheses 1 over straight x subscript 1 close parentheses squared plus open parentheses 1 over straight x subscript 1 close parentheses cubed plus horizontal ellipsis with underbrace below with straight A below plus stack stack open parentheses 1 over straight x subscript 2 close parentheses plus open parentheses 1 over straight x subscript 2 close parentheses squared plus open parentheses 1 over straight x subscript 2 close parentheses cubed plus horizontal ellipsis with underbrace below with straight B below equals...

  • straight A equals open parentheses 1 over straight x subscript 1 close parentheses plus open parentheses 1 over straight x subscript 1 close parentheses squared plus open parentheses 1 over straight x subscript 1 close parentheses cubed plus horizontal ellipsis

open square brackets Jumlah straight space tak straight space hingga straight space dengan straight space straight a equals 1 over straight x subscript 1 straight space dan straight space straight b equals fraction numerator open parentheses 1 over straight x subscript 1 close parentheses squared over denominator open parentheses 1 over straight x subscript 1 close parentheses end fraction close square brackets comma straight S subscript straight infinity equals fraction numerator straight a over denominator 1 minus straight r end fraction space

straight A equals fraction numerator 1 over straight x subscript 1 over denominator 1 minus 1 over straight x subscript 1 end fraction equals 1 over straight x subscript 1. fraction numerator straight x subscript 1 over denominator open parentheses straight x subscript 1 minus 1 close parentheses end fraction equals fraction numerator 1 over denominator straight x subscript 1 minus 1 end fraction

  • straight B equals open parentheses 1 over straight x subscript 2 close parentheses plus open parentheses 1 over straight x subscript 2 close parentheses squared plus open parentheses 1 over straight x subscript 2 close parentheses cubed plus horizontal ellipsis

open square brackets Jumlah straight space tak straight space hingga straight space dengan straight space straight a equals 1 over straight x subscript 2 straight space dan straight space straight b equals fraction numerator open parentheses 1 over straight x subscript 2 close parentheses squared over denominator open parentheses 1 over straight x subscript 2 close parentheses end fraction close square brackets comma space straight S subscript straight infinity equals fraction numerator straight a over denominator 1 minus straight r end fraction  straight B equals fraction numerator 1 over straight x subscript 2 over denominator 1 minus 1 over straight x subscript 2 end fraction equals 1 over straight x subscript 2. fraction numerator straight x subscript 2 over denominator open parentheses straight x subscript 2 minus 1 close parentheses end fraction equals fraction numerator 1 over denominator straight x subscript 2 minus 1 end fraction

 

Jadi jumlah tak hingga di atas adalah

straight A plus straight B equals fraction numerator 1 over denominator straight x subscript 1 minus 1 end fraction plus fraction numerator 1 over denominator straight x subscript 2 minus 1 end fraction equals fraction numerator straight x subscript 2 minus 1 plus straight x subscript 1 minus 1 over denominator open parentheses straight x subscript 1 minus 1 close parentheses open parentheses straight x subscript 2 minus 1 close parentheses end fraction equals fraction numerator straight x subscript 1 plus straight x subscript 2 over denominator straight x subscript 1 straight x subscript 2 minus open parentheses straight x subscript 1 plus straight x subscript 2 close parentheses plus 1 end fraction  equals fraction numerator 8 over denominator 9 minus 8 plus 1 end fraction  equals 8 over 2  equals space 4

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