Giải bài 10 trang 78 SGK giải tích nâng cao 12

a) Ta có: 

$\begin{align} & \sqrt{4+2\sqrt{3}}=\sqrt{{3}+2\sqrt{3}+1} \\ & =\sqrt{{{\left( \sqrt{3}+1 \right)}^{2}}} \\ & =\sqrt{3}+1 \end{align}$

Tương tự $\sqrt{4-2\sqrt{3}}=\sqrt{3}-1$

Suy ra $\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=2$ (đpcm)

b) Ta có

$\begin{align} \sqrt[3]{9+\sqrt{80}}& =\sqrt[3]{9+4\sqrt{5}} \\ & =\sqrt[3]{5+4\sqrt{5}+4} \\ & =\sqrt[3]{{{\left( \sqrt{5}+2 \right)}^{2}}} \end{align}$

Tương tự $\sqrt[3]{9-\sqrt{80}}=\sqrt[3]{{{\left( \sqrt{5}-2 \right)}^{2}}}$

Đặt $\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}=t$

Nhắc lại: ${{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3xy\left( x+y \right)$

Ta có 

${{\left( t \right)}^{3}}=9+\sqrt{80}+9-\sqrt{80}+3t\sqrt[3]{{{\left( \sqrt{5}+2 \right)}^{2}}{{\left( \sqrt{5}-2 \right)}^{2}}} \\ =18+3t\sqrt[3]{{{\left[ {{\left( \sqrt{5} \right)}^{2}}-4 \right]}^{2}}} \\ =18+3t \\ \Leftrightarrow {{t}^{3}}-3t-18=0 \\ \Leftrightarrow \left( t-3 \right)\left( {{t}^{2}}+3t+6 \right)=0 \\ \Leftrightarrow \left[ \begin{align} & t=3 \\ & {{t}^{2}}+3t+6=0\,\left( VN \right) \\ \end{align} \right.$

$t = 3$ hay $\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}=3$ (đpcm)