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Pertanyaan

Jika p = ( x 2 3 ​ + x 2 1 ​ ) ( x 3 1 ​ − x − 3 1 ​ ) dan q = ( x 2 1 ​ + x − 2 1 ​ ) ( x − x 3 1 ​ ) , maka q p ​ = ...

Jika  dan , maka 

  1. ...space 

  2. ...space 

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A. Rahmawati

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah .

jawaban yang tepat adalah begin mathsize 14px style x to the power of 1 third end exponent end style.

Pembahasan

Jadi, jawaban yang tepat adalah .

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell p over q end cell equals cell fraction numerator open parentheses x to the power of begin display style 3 over 2 end style end exponent plus x to the power of begin display style 1 half end style end exponent close parentheses open parentheses x to the power of begin display style 1 third end style end exponent minus x to the power of negative begin display style 1 third end style end exponent close parentheses over denominator open parentheses x to the power of begin display style 1 half end style end exponent plus x to the power of negative begin display style 1 half end style end exponent close parentheses open parentheses x minus x to the power of begin display style 1 third end style end exponent close parentheses end fraction end cell row blank equals cell fraction numerator x to the power of begin display style 1 half end style end exponent open parentheses x plus 1 close parentheses times x to the power of negative begin display style 1 third end style end exponent open parentheses x to the power of begin display style 2 over 3 end style end exponent minus 1 close parentheses over denominator x to the power of negative begin display style 1 half end style end exponent open parentheses x plus 1 close parentheses times x to the power of begin display style 1 third end style end exponent open parentheses x to the power of begin display style 2 over 3 end style end exponent minus 1 close parentheses end fraction end cell row blank equals cell fraction numerator x to the power of begin display style 1 half end style end exponent up diagonal strike open parentheses x plus 1 close parentheses end strike times x to the power of negative begin display style 1 third end style end exponent up diagonal strike open parentheses x to the power of begin display style 2 over 3 end style end exponent minus 1 close parentheses end strike over denominator x to the power of negative begin display style 1 half end style end exponent up diagonal strike open parentheses x plus 1 close parentheses end strike times x to the power of begin display style 1 third end style end exponent up diagonal strike open parentheses x to the power of begin display style 2 over 3 end style end exponent minus 1 close parentheses end strike end fraction end cell row blank equals cell fraction numerator x to the power of begin display style 1 half end style end exponent times x to the power of negative begin display style 1 third end style end exponent over denominator x to the power of negative begin display style 1 half end style end exponent times x to the power of begin display style 1 third end style end exponent end fraction end cell row blank equals cell x to the power of 1 half end exponent times x to the power of negative 1 third end exponent times x to the power of 1 half end exponent times x to the power of negative 1 third end exponent end cell row blank equals cell x to the power of 1 half minus 1 third plus 1 half minus 1 third end exponent end cell row blank equals cell x to the power of 1 minus 2 over 3 end exponent end cell row blank equals cell x to the power of 1 third end exponent end cell end table end style

Jadi, jawaban yang tepat adalah begin mathsize 14px style x to the power of 1 third end exponent end style.

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