Roboguru

Tunjukkan kebenaran pernyataan berikut dengan prinsip induksi matematika.

Pertanyaan

Tunjukkan kebenaran pernyataan berikut dengan prinsip induksi matematika.

begin mathsize 14px style sum from k equals 1 to n of fraction numerator 1 over denominator k open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses end fraction equals 1 half open square brackets 1 half minus fraction numerator 1 over denominator open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses end fraction close square brackets end style

Pembahasan Soal:

Langkah-langkah induksi:

1. Buktikan untuk bilangan 1, pernyataan tersebut benar.

(1)(1+1)(1+2)1(1)(2)(3)16161====21[21(1+1)(1+2)1]21(2161)21×6261 

Untuk n=1 pernyataan benar.

2. Nyatakan untuk bilangan asli sembarang, misalnya n=k, pernyataan tersebut diasumsikan benar.

k=1kk(k+1)(k+2)1=21[21(k+1)(k+2)1] 

3. Untuk n=k+1 akan dibuktikan

k=1k+1k(k+1)(k+2)1=21[21((k+1)+1)((k+1)+2)1] 

Maka:

=======k=1kk(k+1)(k+2)1+(k+1)(k+2)(k+3)121[21(k+1)(k+2)1]+(k+1)(k+2)(k+3)1412(k+1)(k+2)1+(k+1)(k+2)(k+3)1412(k+1)(k+2)(k+3)k+32412(k+1)(k+2)(k+3)k+1412(k+2)(k+3)121[21(k+2)(k+3)1]21[21((k+1)+1)((k+1)+2)1] 

Dengan demikian, Pernyataanbegin mathsize 14px style sum from k equals 1 to n of fraction numerator 1 over denominator k open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses end fraction equals 1 half open square brackets 1 half minus fraction numerator 1 over denominator open parentheses n plus 1 close parentheses open parentheses n plus 2 close parentheses end fraction close square brackets end style merupakan pernyataan yang benar.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 16 September 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

i=1∑n​(2i)=n(n+1)

Pembahasan Soal:

Tidak terdapat perintah dalam soal. Kita asumsikan bahwa perintahnya adalah membuktikan pernyataan pada soal. Kita dapat menggunakan induksi matematika untuk membuktikannya. Langkah-langkah pembuktian menggunakan induksi matematika adalah sebagai berikut:

  • tunjukkan pernyataan benar untuk n=1
  • asumsikan pernyataan benar untuk n=k
  • tunjukkan pernyataan benar untuk n=k+1

Notasi i=1n(2i)=n(n+1) dapat juga dituliskan 2+4+6+...+2n=n(n+1)

  • untuk n=1

2(1)2==1(1+1)2

terbukti benar untuk n=1

  • asumsikan n=k benar

2+4+6+...+2k=k(k+1)

  • untuk n=k+1

2+4+6+...+2k+2(k+1)k(k+1)+2(k+1)k2+k+2k+2k2+3k+2====(k+1)((k+1)+1)(k+1)((k+1)+1)(k+1)(k+2)k2+3k+2

terbukti benar untuk n=k+1

Dengan demikian i=1n(2i)=n(n+1) terbukti benar.

0

Roboguru

Buktikan dengan induksi matematika.

Pembahasan Soal:

Prinsip Induksi Matematika:

Misalkan P open parentheses n close parentheses merupakan suatu pernyataan untuk setiap bilangan asli n. Pernyataan P open parentheses n close parentheses benar jika memenuhi langkah berikut.

1. Langkah awal: Dibuktikan P open parentheses 1 close parentheses benar.

2. Langkah induksi: Jika diasumsikan P open parentheses k close parentheses benar, maka harus dibuktikan bahwa P open parentheses k plus 1 close parentheses juga benar, untuk setiap k bilangan asli.

Jika langkah 1 dan 2 sudah diuji kebenarannya, maka ditarik kesimpulan bahwa P open parentheses n close parentheses benar untuk setiap bilangan asli n.

Akan dibuktikan dengan induksi matematika bahwa

fraction numerator 1 over denominator 1 cross times 2 end fraction plus fraction numerator 1 over denominator 2 cross times 3 end fraction plus horizontal ellipsis plus fraction numerator 1 over denominator n open parentheses n plus 1 close parentheses end fraction equals fraction numerator n over denominator n plus 1 end fraction

Langkah awal:

Akan dibuktikan P open parentheses n close parentheses benar untuk n equals 1

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 over denominator 1 open parentheses 1 plus 1 close parentheses end fraction end cell equals cell fraction numerator 1 over denominator 1 plus 1 end fraction end cell row cell 1 half end cell equals cell 1 half end cell end table

Jadi, terbukti bahwa P open parentheses 1 close parentheses benar.

Langkah induksi:

Asumsikan P open parentheses k close parentheses benar sehingga diperoleh

fraction numerator 1 over denominator 1 cross times 2 end fraction plus fraction numerator 1 over denominator 2 cross times 3 end fraction plus horizontal ellipsis plus fraction numerator 1 over denominator k open parentheses k plus 1 close parentheses end fraction equals fraction numerator k over denominator k plus 1 end fraction

Akan ditunjukkan bahwa P open parentheses k plus 1 close parentheses juga benar, sedemikian sehingga 

fraction numerator 1 over denominator 1 cross times 2 end fraction plus fraction numerator 1 over denominator 2 cross times 3 end fraction plus horizontal ellipsis plus fraction numerator 1 over denominator k open parentheses k plus 1 close parentheses end fraction plus fraction numerator 1 over denominator open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses end fraction equals fraction numerator k plus 1 over denominator open parentheses k plus 1 close parentheses plus 1 end fraction

Bukti:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 1 over denominator 1 cross times 2 end fraction plus fraction numerator 1 over denominator 2 cross times 3 end fraction plus horizontal ellipsis plus fraction numerator 1 over denominator k open parentheses k plus 1 close parentheses end fraction plus fraction numerator 1 over denominator open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses end fraction end cell row blank equals cell fraction numerator k over denominator k plus 1 end fraction plus fraction numerator 1 over denominator open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses end fraction end cell row blank equals cell fraction numerator k open parentheses k plus 2 close parentheses plus 1 over denominator open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses end fraction end cell row blank equals cell fraction numerator k squared plus 2 k plus 1 over denominator open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses k plus 1 close parentheses open parentheses k plus 1 close parentheses over denominator open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses end fraction end cell row blank equals cell fraction numerator k plus 1 over denominator k plus 2 end fraction end cell row blank equals cell fraction numerator k plus 1 over denominator open parentheses k plus 1 close parentheses plus 1 end fraction end cell end table

Jadi, terbukti bahwa P open parentheses k plus 1 close parentheses benar .

Pernyataan P open parentheses n close parentheses memenuhi kedua prinsip induksi matematika.

Dengan demikian, berdasarkan prinsip induksi matematika, P open parentheses n close parentheses benar untuk setiap n bilangan asli.

3

Roboguru

Diberikan pernyataan sebagai berikut. untuk setiap bilangan asli . Dengan menggunakan induksi matematika, dapat disimpulkan bahwa ....

Pembahasan Soal:

Diberikan pernyataan sebagai berikut.

P subscript n colon sum from i equals 1 to n of fraction numerator i plus 2 over denominator i times open parentheses i plus 1 close parentheses times 2 to the power of i end fraction equals 1 minus fraction numerator 1 over denominator open parentheses n plus 1 close parentheses times 2 to the power of n end fraction

untuk setiap bilangan asli n.

Karena akan dibuktikan pernyataan untuk setiap bilangan asli n, yaitu begin mathsize 14px style n greater or equal than 1 end style, maka langkah pertamanya adalah buktikan P subscript 1 benar.

LANGKAH 1: Buktikan P subscript 1 benar.

Perhatikan pernyataan berikut.

P subscript n colon sum from i equals 1 to n of fraction numerator i plus 2 over denominator i times open parentheses i plus 1 close parentheses times 2 to the power of i end fraction equals 1 minus fraction numerator 1 over denominator open parentheses n plus 1 close parentheses times 2 to the power of n end fraction

Substitusikan nilai n equals 1 seperti berikut ini.

P subscript 1 colon sum from i equals 1 to 1 of fraction numerator i plus 2 over denominator i times open parentheses i plus 1 close parentheses times 2 to the power of i end fraction equals 1 minus fraction numerator 1 over denominator open parentheses 1 plus 1 close parentheses times 2 to the power of 1 end fraction

Dari ruas kiri, didapatkan perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell sum from i equals 1 to 1 of fraction numerator i plus 2 over denominator i times open parentheses i plus 1 close parentheses times 2 to the power of i end fraction end cell equals cell fraction numerator 1 plus 2 over denominator 1 times open parentheses 1 plus 1 close parentheses times 2 to the power of 1 end fraction end cell row blank equals cell fraction numerator 3 over denominator 1 times 2 times 2 end fraction end cell row blank equals cell 3 over 4 end cell end table

Dari ruas kanan, didapat perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 1 minus fraction numerator 1 over denominator open parentheses 1 plus 1 close parentheses times 2 to the power of 1 end fraction end cell equals cell 1 minus fraction numerator 1 over denominator 2 times 2 end fraction end cell row blank equals cell 1 minus 1 fourth end cell row blank equals cell 3 over 4 end cell end table end style

Karena ruas kiri sama dengan ruas kanan, maka P subscript 1 benar.

LANGKAH 2: Buktikan untuk sembarang bilangan asli k , P subscript k bernilai benar mengakibatkan P subscript k plus 1 end subscript bernilai benar.

Perhatikan pernyataan berikut.

P subscript n colon sum from i equals 1 to n of fraction numerator i plus 2 over denominator i times open parentheses i plus 1 close parentheses times 2 to the power of i end fraction equals 1 minus fraction numerator 1 over denominator open parentheses n plus 1 close parentheses times 2 to the power of n end fraction

Asumsikan pernyataan tersebut benar untuk sembarang bilangan asli k seperti berikut ini.

P subscript k colon sum from i equals 1 to k of fraction numerator i plus 2 over denominator i times open parentheses i plus 1 close parentheses times 2 to the power of i end fraction equals 1 minus fraction numerator 1 over denominator open parentheses k plus 1 close parentheses times 2 to the power of k end fraction

Substitusikan nilai n equals k plus 1 sebagai berikut.

P subscript k plus 1 end subscript colon sum from i equals 1 to k plus 1 of fraction numerator i plus 2 over denominator i times open parentheses i plus 1 close parentheses times 2 to the power of i end fraction equals 1 minus fraction numerator 1 over denominator open parentheses open parentheses k plus 1 close parentheses plus 1 close parentheses times 2 to the power of k plus 1 end exponent end fraction

Dari ruas kiri P subscript k plus 1 end subscript didapat perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sum from i equals 1 to k plus 1 of fraction numerator i plus 2 over denominator i times open parentheses i plus 1 close parentheses times 2 to the power of i end fraction end cell row blank equals cell sum from i equals 1 to k of fraction numerator i plus 2 over denominator i times open parentheses i plus 1 close parentheses times 2 to the power of i end fraction plus fraction numerator open parentheses k plus 1 close parentheses plus 2 over denominator open parentheses k plus 1 close parentheses times open parentheses open parentheses k plus 1 close parentheses plus 1 close parentheses times 2 to the power of open parentheses k plus 1 close parentheses end exponent end fraction end cell row blank equals cell open parentheses 1 minus fraction numerator 1 over denominator open parentheses k plus 1 close parentheses times 2 to the power of k end fraction close parentheses plus fraction numerator open parentheses k plus 1 close parentheses plus 2 over denominator open parentheses k plus 1 close parentheses times open parentheses open parentheses k plus 1 close parentheses plus 1 close parentheses times 2 to the power of open parentheses k plus 1 close parentheses end exponent end fraction end cell row blank equals cell 1 minus fraction numerator 1 over denominator open parentheses k plus 1 close parentheses times 2 to the power of k end fraction plus fraction numerator k plus 3 over denominator open parentheses k plus 1 close parentheses times open parentheses k plus 2 close parentheses times 2 to the power of k times 2 to the power of 1 end fraction end cell row blank equals cell 1 minus fraction numerator 1 over denominator open parentheses k plus 1 close parentheses times 2 to the power of k end fraction times open parentheses 1 minus fraction numerator k plus 3 over denominator open parentheses k plus 2 close parentheses times 2 end fraction close parentheses end cell row blank equals cell 1 minus fraction numerator 1 over denominator open parentheses k plus 1 close parentheses times 2 to the power of k end fraction times open parentheses 1 minus fraction numerator k plus 3 over denominator 2 k plus 4 end fraction close parentheses end cell row blank equals cell 1 minus fraction numerator 1 over denominator open parentheses k plus 1 close parentheses times 2 to the power of k end fraction times open parentheses fraction numerator 2 k plus 4 over denominator 2 k plus 4 end fraction minus fraction numerator k plus 3 over denominator 2 k plus 4 end fraction close parentheses end cell row blank equals cell 1 minus fraction numerator 1 over denominator open parentheses k plus 1 close parentheses times 2 to the power of k end fraction times fraction numerator open parentheses 2 k plus 4 close parentheses minus open parentheses k plus 3 close parentheses over denominator 2 k plus 4 end fraction end cell row blank equals cell 1 minus fraction numerator 1 over denominator open parentheses k plus 1 close parentheses times 2 to the power of k end fraction times fraction numerator k plus 1 over denominator 2 open parentheses k plus 2 close parentheses end fraction end cell row blank equals cell 1 minus fraction numerator 1 over denominator up diagonal strike open parentheses k plus 1 close parentheses end strike times 2 to the power of k end fraction times fraction numerator up diagonal strike k plus 1 end strike over denominator 2 open parentheses k plus 2 close parentheses end fraction end cell row blank equals cell 1 minus 1 over 2 to the power of k times fraction numerator 1 over denominator 2 open parentheses k plus 2 close parentheses end fraction end cell row blank equals cell 1 minus fraction numerator 1 over denominator open parentheses k plus 2 close parentheses times 2 to the power of k plus 1 end exponent end fraction end cell row blank equals cell 1 minus fraction numerator 1 over denominator open parentheses open parentheses k plus 1 close parentheses plus 1 close parentheses times 2 to the power of k plus 1 end exponent end fraction end cell end table

Dapat diperhatikan bahwa ruas kiri sama dengan ruas kanan. Jadi, P subscript k plus 1 end subscript bernilai benar.

Dari penjabaran di atas, didapatkan informasi berikut.

  1. P subscript 1 benar.
  2. Untuk sembarang bilangan asli k, jika P subscript k bernilai benar mengakibatkan P subscript k plus 1 end subscript bernilai benar.

Akibatnya, P subscript n benar untuk setiap bilangan asli n, menurut prinsip induksi matematika.

Dengan demikian, pada proses pembuktian dengan induksi matematika di atas, didapatkan bahwa pernyataan terbukti dengan induksi matematika.

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Buktikanlah. c.

Pembahasan Soal:

Membuktikan dengan menggunakan induksi matematika dimana

Untuk n =1

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus straight n times 2 to the power of straight n minus 1 end exponent end cell equals cell left parenthesis straight n minus 1 right parenthesis times 2 to the power of straight n plus 1 end cell row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus 1 times 2 to the power of 1 minus 1 end exponent end cell equals cell left parenthesis 1 minus 1 right parenthesis times 2 to the power of 1 plus 1 end cell row cell 1.2 to the power of 0 end cell equals cell 0.2 plus 1 end cell row 1 equals cell 1 rightwards arrow terbukti end cell end table

Untuk n =k diasumsikan terbukti 

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus straight n times 2 to the power of straight n minus 1 end exponent end cell equals cell left parenthesis straight n minus 1 right parenthesis times 2 to the power of straight n plus 1 end cell row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus straight k times 2 to the power of straight k minus 1 end exponent end cell equals cell left parenthesis straight k minus 1 right parenthesis times 2 to the power of straight k plus 1 rightwards arrow terbukti end cell end table

Untuk n = k+1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus straight n times 2 to the power of straight n minus 1 end exponent end cell equals cell left parenthesis straight n minus 1 right parenthesis times 2 to the power of straight n plus 1 end cell row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 plus 1 end exponent end cell equals cell left parenthesis straight k plus 1 minus 1 right parenthesis times 2 to the power of straight k plus 1 end exponent plus 1 end cell row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus open parentheses straight k plus 1 close parentheses times 2 to the power of straight k end cell equals cell left parenthesis straight k right parenthesis times 2 to the power of straight k plus 1 end exponent plus 1 end cell row cell 2 straight k to the power of straight k plus 2 to the power of straight k end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 1 rightwards arrow terbukti end cell end table

Maka akan dibuktikan

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus straight n times 2 to the power of straight n minus 1 end exponent end cell equals cell left parenthesis straight n minus 1 right parenthesis times 2 to the power of straight n plus 1 end cell row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus k.2 to the power of k plus open parentheses straight k plus 1 close parentheses times 2 to the power of straight k minus 1 plus 1 end exponent end cell equals cell left parenthesis straight k plus 1 minus 1 right parenthesis times 2 to the power of straight k plus 1 end exponent plus 1 end cell row cell 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus 1 left parenthesis straight k minus 1 right parenthesis.2 to the power of straight k plus 2 straight k to the power of straight k plus 2 to the power of straight k end cell equals cell left parenthesis straight k right parenthesis times 2 to the power of straight k plus 1 end exponent plus 1 end cell row cell 1 plus 2 straight k to the power of straight k minus 2 to the power of straight k plus 2 straight k to the power of straight k plus 2 to the power of straight k end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell row cell 1 plus 4 straight k to the power of straight k end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell row cell 1 plus 2 straight k.2 to the power of straight k end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell row cell 1 plus straight k.2 to the power of 1.2 to the power of straight k end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell row cell 2 straight k to the power of straight k plus 1 end exponent plus 1 end cell equals cell 2 straight k to the power of straight k plus 1 end exponent plus 1 rightwards arrow terbukti end cell end table

Jadi terbukti 1 times 2 to the power of 0 plus 2 times 2 to the power of 1 plus 3 times 2 squared plus 4 times 2 cubed plus.... plus straight n times 2 to the power of straight n minus 1 end exponent equals left parenthesis straight n minus 1 right parenthesis times 2 to the power of straight n plus 1 karena hasil sisi kanan dan kiri sama

 

0

Roboguru

Buktikanlah. b.

Pembahasan Soal:

Pembuktian menggunakan induksi matematika

Dimana untuk n = 1

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of straight n end cell equals cell 1 half open parentheses 3 to the power of straight n plus 1 end exponent minus 3 close parentheses end cell row cell 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of 1 end cell equals cell 1 half open parentheses 3 to the power of 1 plus 1 end exponent minus 3 close parentheses end cell row 3 equals cell 1 half open parentheses 3 squared minus 3 close parentheses end cell row 3 equals cell 1 half open parentheses 6 close parentheses end cell row 3 equals cell 3 rightwards arrow Terbukti space end cell end table

Untuk n = k diasumsikan terbukti maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of straight n end cell equals cell 1 half open parentheses 3 to the power of straight n plus 1 end exponent minus 3 close parentheses end cell row cell 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of straight k end cell equals cell 1 half open parentheses 3 to the power of straight k plus 1 end exponent minus 3 close parentheses end cell row cell 3 to the power of straight k end cell equals cell 1 half open parentheses 3 to the power of straight k plus 1 end exponent minus 3 close parentheses end cell row cell 3 to the power of straight k end cell equals cell 1 half open parentheses 3 to the power of straight k plus 1 end exponent minus 3 close parentheses rightwards arrow Terbukti space end cell end table

Untuk n = k+1 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of straight n end cell equals cell 1 half open parentheses 3 to the power of straight n plus 1 end exponent minus 3 close parentheses end cell row cell 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of straight k plus 3 to the power of straight k plus 1 end exponent end cell equals cell 1 half open parentheses 3 to the power of straight k plus 2 end exponent minus 3 close parentheses end cell row cell 1 half open parentheses 3 to the power of straight k plus 1 end exponent minus 3 close parentheses plus 3 to the power of straight k plus 1 end exponent end cell equals cell 1 half open parentheses 3 to the power of straight k plus 2 end exponent minus 3 close parentheses end cell row cell 1 half open parentheses 3 to the power of straight k plus 2 end exponent minus 3 close parentheses end cell equals cell 1 half open parentheses 3 to the power of straight k plus 2 end exponent minus 3 close parentheses rightwards arrow Terbukti space end cell end table

Jadi terbukti bahwa 3 plus 3 squared plus 3 cubed plus... plus 3 to the power of straight n equals 1 half open parentheses 3 to the power of straight n plus 1 end exponent minus 3 close parentheses karena hasil dari sisi kanan dan kiri sama

 

1

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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