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x3y−2xy5​=....

Pertanyaan

x3y2xy5=.... 

Pembahasan Soal:

Untuk pertidaksamaan bentuk f(x)g(x) maka ruas kiri sudah pasti bernilai positif namun ruas kanan belum tentu bernilai positif, sehingga bentuk pertidaksamaan seperti ini perlu diuraikan menjadi dua kemungkinan yaitu g(x)0dang(x)<0

Kemungkinan Kasus 1 : x3<0x>3

Olehkarena3x+10danx<3maka3x+1x3terpenuhi untuk semua x.

Syarat akar : 

 3x+1x031

Irisan dari x<3,xR,danx31 direpresentasikan oleh garais bilangan berikut :

Sehingga HP kasus 1 :31x<3 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Septianingsih

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 30 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai  yang memenuhi penyelesaian dari  adalah ....

Pembahasan Soal:

Error converting from MathML to accessible text.

abaikan angka 2 sehingga fokus pada pangkat.

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Jadi, jawaban yang tepat adalah A.

0

Roboguru

Bentuk sederhana dari  adalah ....

Pembahasan Soal:

Ingat kembali konsep bilangan berpangkat.

table attributes columnalign right center left columnspacing 2px 2px 2px end attributes row cell open parentheses a to the power of b close parentheses to the power of c end cell equals cell a to the power of b cross times c end exponent end cell row cell open parentheses a cross times b close parentheses to the power of m end cell equals cell a to the power of m cross times b to the power of m end cell row cell a to the power of m over a to the power of n end cell equals cell a to the power of m minus n end exponent end cell row cell a to the power of negative n end exponent end cell equals cell 1 over a to the power of n end cell end table

Akan ditentukan bentuk sederhana dari open parentheses a squared b to the power of negative 2 end exponent close parentheses to the power of 5 over open parentheses open parentheses 2 a close parentheses cubed b squared close parentheses to the power of negative 2 end exponent.

table attributes columnalign right center left columnspacing 2px end attributes row cell open parentheses a squared b to the power of negative 2 end exponent close parentheses to the power of 5 over open parentheses open parentheses 2 a close parentheses cubed b squared close parentheses to the power of negative 2 end exponent end cell equals cell fraction numerator open parentheses a squared close parentheses to the power of 5 open parentheses b to the power of negative 2 end exponent close parentheses to the power of 5 over denominator open parentheses 8 a cubed b squared close parentheses to the power of negative 2 end exponent end fraction end cell row blank equals cell fraction numerator a to the power of 2 cross times 5 end exponent b to the power of negative 2 cross times 5 end exponent over denominator open parentheses 8 a cubed close parentheses to the power of negative 2 end exponent open parentheses b squared close parentheses to the power of negative 2 end exponent end fraction end cell row blank equals cell fraction numerator a to the power of 10 b to the power of negative 10 end exponent over denominator 8 to the power of negative 2 end exponent a to the power of 3 cross times open parentheses negative 2 close parentheses end exponent b to the power of 2 cross times open parentheses negative 2 close parentheses end exponent end fraction end cell row blank equals cell fraction numerator 8 squared a to the power of 10 b to the power of negative 10 end exponent over denominator a to the power of negative 6 end exponent b to the power of negative 4 end exponent end fraction end cell row blank equals cell 64 times a to the power of 10 over a to the power of negative 6 end exponent times b to the power of negative 10 end exponent over b to the power of negative 4 end exponent end cell row blank equals cell 64 a to the power of 10 minus open parentheses negative 6 close parentheses end exponent b to the power of negative 10 minus open parentheses negative 4 close parentheses end exponent end cell row blank equals cell 64 a to the power of 10 plus 6 end exponent b to the power of negative 10 plus 4 end exponent end cell row blank equals cell 64 a to the power of 16 b to the power of negative 6 end exponent end cell row blank equals cell fraction numerator 64 a to the power of 16 over denominator b to the power of 6 end fraction end cell end table

Diperoleh open parentheses a squared b to the power of negative 2 end exponent close parentheses to the power of 5 over open parentheses open parentheses 2 a close parentheses cubed b squared close parentheses to the power of negative 2 end exponent equals fraction numerator 64 a to the power of 16 over denominator b to the power of 6 end fraction.

Jadi, jawaban yang tepat adalah A.

1

Roboguru

Tentukan hasil dari .

Pembahasan Soal:

Soal ini dapat diselesaikan dengan menggunakan sifat eksponen sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 to the power of 4 cross times 3 to the power of 5 cross times 4 cubed over denominator 12 squared end fraction end cell equals cell fraction numerator 2 to the power of 4 cross times 3 to the power of 5 cross times open parentheses 2 squared close parentheses cubed over denominator open parentheses 3 cross times 4 close parentheses squared end fraction end cell row blank equals cell fraction numerator 2 to the power of 4 cross times 3 to the power of 5 cross times 2 to the power of 6 over denominator 3 squared cross times 4 squared end fraction end cell row blank equals cell fraction numerator 2 to the power of 4 plus 6 end exponent cross times 3 to the power of 5 over denominator 3 squared cross times 2 to the power of 4 end fraction end cell row blank equals cell fraction numerator 2 to the power of 10 cross times 3 to the power of 5 over denominator 2 to the power of 4 cross times 3 squared end fraction end cell row blank equals cell 2 to the power of 10 minus 4 end exponent cross times 3 to the power of 5 minus 2 end exponent end cell row blank equals cell 2 to the power of 6 cross times 3 cubed end cell row blank equals cell 64 cross times 27 end cell row blank equals cell 1.728 end cell end table end style

Sehingga, hasil dari fraction numerator bold 2 to the power of bold 4 bold cross times bold 3 to the power of bold 5 bold cross times bold 4 to the power of bold 3 over denominator bold 12 to the power of bold 2 end fraction bold equals bold 1 bold. bold 728.

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Bentuk sederhana

Pembahasan Soal:

fraction numerator p squared q to the power of 4 horizontal strike r to the power of negative 2 end exponent end strike over denominator p to the power of 5 q to the power of negative 2 end exponent horizontal strike r to the power of negative 2 end exponent end strike end fraction equals fraction numerator q to the power of 4 q squared over denominator p to the power of 5 p to the power of negative 2 end exponent end fraction  space space space space space space space space space space space space space space space space space space space equals q to the power of 4 plus 2 end exponent over p to the power of 5 minus 2 end exponent space space space space space space space space  space space space space space space space space space space space space space space space space space space space equals q to the power of 6 over p cubed

0

Roboguru

Jika , nilai dari

Pembahasan Soal:

Dengan menerapkan sifat bilangan berpangkat, maka dapat dilakukan perhitungan seperti di bawah ini

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator begin mathsize 14px style left parenthesis 3 p squared right parenthesis squared plus left parenthesis 4 p squared right parenthesis squared end style over denominator begin mathsize 14px style left parenthesis 3 p right parenthesis squared minus left parenthesis 2 p right parenthesis squared end style end fraction end cell size 14px equals size 14px 80 row cell fraction numerator begin mathsize 14px style 9 p to the power of 4 plus 16 p to the power of 4 end style over denominator begin mathsize 14px style 9 p squared minus 4 p squared end style end fraction end cell size 14px equals size 14px 80 row cell fraction numerator begin mathsize 14px style 25 p to the power of 4 end style over denominator begin mathsize 14px style 5 p squared end style end fraction end cell size 14px equals size 14px 80 row cell size 14px 5 size 14px p to the power of size 14px 2 end cell size 14px equals size 14px 80 row cell size 14px p to the power of size 14px 2 end cell size 14px equals size 14px 16 end table

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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