Bài 54 (Trang 30 SGK Toán 9, Tập 1):

Rút gọn các biểu thức sau (giả thiết các biểu thức chữ đều có nghĩa):

$\frac{2+\sqrt{2}}{1+\sqrt{2}}; \frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}; \frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}; \frac{\mathrm{a}-\sqrt{\mathrm{a}}}{1-\sqrt{\mathrm{a}}}; \frac{\mathrm{p}-2\sqrt{\mathrm{p}}}{\sqrt{\mathrm{p}}-2}.$

Hướng dẫn giải:

$\frac{2+\sqrt{2}}{1+\sqrt{2}} = \frac{{\left(\sqrt{2}\right)}^2+\sqrt{2}}{1+\sqrt{2}} = \frac{\sqrt{2}(\sqrt{2}+1)}{1+\sqrt{2}} = \sqrt{2};$

Cách khác: $\frac{2+\sqrt{2}}{1+\sqrt{2}} = \frac{(2+\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})} = \frac{2-2\sqrt{2}+\sqrt{2}-{\left(\sqrt{2}\right)}^2}{1-{\left(\sqrt{2}\right)}^2} = \frac{2-\sqrt{2}-2}{1-2}=\frac{-\sqrt{2}}{-1} = \sqrt{2}.$                

$\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}} = \frac{\sqrt{3.5}-\sqrt{5}}{1-\sqrt{3}} = \frac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}} = -\sqrt{5};$

$\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2} = \frac{\sqrt{2^2.3}-\sqrt{6}}{\sqrt{4.2}-2} = \frac{\sqrt{2}.6-\sqrt{6}}{2\sqrt{2}-2} = \frac{\sqrt{6}(\sqrt{2}-1)}{2(\sqrt{2}-2)} = \frac{\sqrt{6}}{2};$

$\frac{\mathrm{a}-\sqrt{\mathrm{a}}}{1-\sqrt{\mathrm{a}}} = \frac{{\left(\sqrt{\mathrm{a}}\right)}^2-\sqrt{\mathrm{a}}}{1-\sqrt{\mathrm{a}}} = \frac{\sqrt{\mathrm{a}}(\sqrt{\mathrm{a}}-1)}{1-\sqrt{\mathrm{a}}} = -\sqrt{\mathrm{a}};$

$\frac{\mathrm{p}-2\sqrt{\mathrm{p}}}{\sqrt{\mathrm{p}}-2} = \frac{{\left(\sqrt{\mathrm{p}}\right)}^2-2\sqrt{\mathrm{p}}}{\sqrt{\mathrm{p}}-2} = \frac{\sqrt{\mathrm{p}}(\sqrt{\mathrm{p}}-2)}{\sqrt{\mathrm{p}}-2} = \sqrt{\mathrm{p}}.$