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[tex]\sqrt[3]{3}+\sqrt{2}\ \textgreater \ \sqrt[3]{9}\ \textgreater \ \sqrt[4]{13}[/tex]
[tex]\;[/tex]
[tex]\Large\textrm{Note}[/tex]
1. Use of calculator is not allowed.
2. Use of approximated value is not allowed.​

Respuesta :

mhanifa

Answer:

  • See below

Step-by-step explanation:

Use same power to compare the values

Prove the first part:

  • [tex]\sqrt[3]{3}+\sqrt{2} >\sqrt[3]{9}[/tex]
  • [tex]\sqrt[6]{3^2}+\sqrt[6]{2^3} >\sqrt[6]{9^2}[/tex]
  • [tex]\sqrt[6]{9}+\sqrt[6]{8} >\sqrt[6]{81}[/tex]
  • [tex]\sqrt[6]{9}+\sqrt[6]{8}> \sqrt[6]{8}+\sqrt[6]{8}=2\sqrt[6]{8}=\sqrt[6]{8*64} =\sqrt[6]{512} >\sqrt[6]{81}[/tex]
  • [tex]512 > 81[/tex]

Prove the second part:

  • [tex]\sqrt[3]{9} >\sqrt[4]{13}[/tex]
  • [tex]\sqrt[12]{9^4} >\sqrt[12]{13^3}[/tex]
  • [tex]\sqrt[12]{6561} >\sqrt[12]{2197}[/tex]
  • [tex]6561 > 2197[/tex]

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