Matematika

Pertanyaan

berapa [tex]\sqrt[3]{2- \sqrt{5} } [/tex] + [tex]\sqrt[3]{2+ \sqrt{5} } [/tex]

1 Jawaban

  • [tex]\\ $Anggap:$ \\ a=\sqrt[3]{2-\sqrt{5}} \\ b=\sqrt[3]{2+\sqrt{5}} \\ a^3+b^3=(a+b)(a^2-ab+b^2) \\ \displaystyle a+b=\frac{a^3-b^3}{a^2-ab+b^2} \\ a+b=\frac{2-\sqrt{5}+2+\sqrt{5}}{\sqrt[3]{(2-\sqrt{5})^2}-\sqrt[3]{(2-\sqrt{5})(2+\sqrt{5})}+\sqrt[3]{(2+\sqrt{5})^2}} \\ \\ a+b=\frac{4}{\sqrt[3]{9-4\sqrt{5}}-\sqrt[3]{-1}+\sqrt[3]{9+4\sqrt{5}}} \\ \\ a+b=\frac{4}{1+\frac{1}{2}(3-\sqrt{5})+\frac{1}{2}(3+\sqrt{5})} \\ \\ a+b=\frac{4}{1+\frac{3}{2}+\frac{3}{2}} \\ \\ a+b=\frac{4}{4} \\ a+b=1[/tex]

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