Giải câu 1 trang 28 sách phát triển năng lực toán 9 tập 1.
a, $\frac{2-\sqrt{3}}{2\sqrt{3}}$ = $\frac{(2-\sqrt{3}).\sqrt{3}}{2.\sqrt{3}.\sqrt{3}}$ = $\frac{2\sqrt{3}-3}{6}$
b, $\frac{2}{\sqrt{2}+\sqrt{3}}$ = $\frac{2.(\sqrt{3}-\sqrt{2})}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}$ = $\frac{2.(\sqrt{3}-\sqrt{2})}{3-2}$ = $2.(\sqrt{3}-\sqrt{2})$
c, $\frac{1}{2-\sqrt{5}}$ = $\frac{1.(2+\sqrt{5})}{(2-\sqrt{5})(2-\sqrt{5})}$ = $\frac{2+\sqrt{5}}{4-5}$ = $-(2+\sqrt{5})$
d, $\frac{6}{\sqrt{2}+\sqrt{2-\sqrt{2}}}$ = $\frac{6.(\sqrt{2}-\sqrt{2-\sqrt{2}})}{(\sqrt{2}+\sqrt{2-\sqrt{2}})(\sqrt{2}-\sqrt{2-\sqrt{2}})}$
= $\frac{6.(\sqrt{2}-\sqrt{2-\sqrt{2}})}{2-2+\sqrt{2}}$ = $\frac{6.(\sqrt{2}-\sqrt{2-\sqrt{2}})}{\sqrt{2}}$ = $\frac{6.\sqrt{2}.(\sqrt{2}-\sqrt{2-\sqrt{2}})}{\sqrt{2}.\sqrt{2}}$
= $\frac{6.(2-\sqrt{2.(2-\sqrt{2})}}{2}$ = $3.(2-\sqrt{4-2\sqrt{2}})$