Giải câu 5 trang 37 sách phát triển năng lực toán 9 tập 1

a, $(1+\sqrt{2})^{3}+(1-\sqrt{2})^{3}$ = $1+3.1.\sqrt{2}+3.1.2+(\sqrt{2})^{3}+1-3.1\sqrt{2}+3.1.2-(\sqrt{2})^{3}$ = 1 + 6 + 1 + 6 = 14

b, $\sqrt[3]{10+6\sqrt{3}}$ = $\sqrt[3]{1+3\sqrt{3}+9+3\sqrt{3}}$ = $\sqrt[3]{1^{3}+3.1^{2}.\sqrt{3}+3.1.(\sqrt{3})^{2}+(\sqrt{3})^{3}}$

= $\sqrt[3]{(1+\sqrt{3})^{3}}$ = $1+\sqrt{3}$

c, $\frac{4+2\sqrt{3}}{\sqrt[3]{10+6\sqrt{3}}}$ = $\frac{1^{2}+2\sqrt{3}+(\sqrt{3})^{2}}{\sqrt[3]{(1+\sqrt{3})^{3}}}$ = $\frac{(1+\sqrt{3})^{2}}{1+\sqrt{3}}$ = $1+\sqrt{3}$

d, $\sqrt{3+\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}}$ = $\sqrt{3+\sqrt{3}+\sqrt[3]{(1+\sqrt{3})^{3}}}$ = $\sqrt{3+\sqrt{3}+1+\sqrt{3}}$

 = $\sqrt{1^{2}+2\sqrt{3}+(\sqrt{3})^{2}}$ = $\sqrt{(1+\sqrt{3})^{2}}$ = $1+\sqrt{3}$