Question

Mostre que: $\sqrt{4+3\sqrt{2} } =\sqrt[4]{2}+\sqrt[4]{8}$

Answer 1

Explicação passo a passo:

Vamos elevar o lado direito ao quadrado:

    $(\sqrt[4]{2}+\sqrt[4]{8})^2\\\\=(\sqrt[4]{2})^2+2\cdot \sqrt[4]{2}\cdot \sqrt[4]{8}+(\sqrt[4]{8})^2\\\\=\sqrt[4]{2^2}+2\cdot \sqrt[4]{2\cdot 8}+\sqrt[4]{8^2}\\\\=2^{2/4}+2\cdot \sqrt[4]{16}+8^{2/4}$

    $=2^{1/2}+2\cdot \sqrt[4]{2^4}+(2^3)^{1/2}\\\\=2^{1/2}+2\cdot 2+2^{3/2}\\\\=\sqrt{2}+4+\sqrt{2^3}\\\\=\sqrt{2}+4+\sqrt{2^2\cdot 2}$

    $=\sqrt{2}+4+2\sqrt{2}\\\\=4+3\sqrt{2}$

    $\therefore\quad (\sqrt[4]{2}+\sqrt[4]{8})^2=4+3\sqrt{2}\\\\\Longrightarrow\quad \sqrt{4+3\sqrt{2}}=\sqrt[4]{2}+\sqrt[4]{8}\qquad \blacksquare$

Dúvidas? Comente.

Bons estudos!