Cho a ≥ 0, a ≠ 1. Rút gọn biểu thức sau
$S = \sqrt {6 – 4\sqrt 2 } .\sqrt[3]{{20 + 14\sqrt 2 }} + \sqrt[3]{{\left( {a + 3} \right)\sqrt a – 3a – 1}}:\left[ {\frac{{a – 1}}{{2\left( {\sqrt a – 1} \right)}} – 1} \right] + 2019$
Cho a ≥ 0, a ≠ 1. Rút gọn biểu thức sau
$S = \sqrt {6 – 4\sqrt 2 } .\sqrt[3]{{20 + 14\sqrt 2 }} + \sqrt[3]{{\left( {a + 3} \right)\sqrt a – 3a – 1}}:\left[ {\frac{{a – 1}}{{2\left( {\sqrt a – 1} \right)}} – 1} \right] + 2019$
$6-4\sqrt{2}=4+2+2.2.\sqrt{2}=(2-\sqrt{2})^{2}$ ⇒ $\sqrt{6-4\sqrt{2}}=\sqrt{(2-\sqrt{2})^{2}}=|2-\sqrt{2}|=2-\sqrt{2}$
$20+14\sqrt{2}=8+12+2\sqrt{2}+12\sqrt{2}=2^{3}+3.2^{2}\sqrt{2}+3.2.\sqrt{2}^{2}+\sqrt{2^{3}}=(2+\sqrt{2})^{3}$ suy ra $\sqrt[3]{20+14\sqrt{2}}=\sqrt[3]{(2+\sqrt{2})^{3}}=2+\sqrt{2}$
$\sqrt[3]{(a+3)\sqrt{a}-3a-1}=\sqrt[3]{a.\sqrt{a}+3\sqrt{a}-3a-1}=\sqrt[3]{\sqrt{1^{3}}-3.1^{2}\sqrt{a}+3.1.\sqrt{a}+\sqrt{a^{3}}}=\sqrt[3]{(1-\sqrt{a})^{3}}=1-\sqrt{a}$
Và $\frac{a-1}{2(\sqrt{a}-1)}-1=\frac{(\sqrt{a}-1)(\sqrt{a}+1)}{2(\sqrt{a}-1)}-1=\frac{\sqrt{a}+1}{2}-1=\frac{\sqrt{a}-1}{2}$
Vậy $(2+\sqrt{2})(2-\sqrt{2})+(1-\sqrt{a})/\frac{\sqrt{a}-1}{2}+2019=2^{2}-\sqrt{2^{2}}-2+2019=4-2-2+2019=2019$