Bài 49 (Trang 29 SGK Toán 9, Tập 1):

Khử mẫu của biểu thức lấy căn:

$ab\sqrt{\frac{a}{b}}; \frac{a}{b}\sqrt{\frac{b}{a}}; \sqrt{\frac{1}{b}+\frac{1}{b^2}}; \sqrt{\frac{9a^3}{36b}}; 3xy\sqrt{\frac{2}{xy}}.$

(Giả thiết các biểu thức có nghĩa).

Hướng dẫn giải:

$\mathrm{ab}\sqrt{\frac{\mathrm{a}}{\mathrm{b}}} = \mathrm{ab}\sqrt{\frac{\mathrm{ab}}{{\mathrm{b}}^2}} = \frac{\mathrm{ab}}{\left|\mathrm{b}\right|}\sqrt{\mathrm{ab}} = \left\{\begin{array}{l}\mathrm{a}\sqrt{\mathrm{ab}} \mathrm{nếu} \mathrm{b} > 0\\ -\mathrm{a}\sqrt{\mathrm{ab}} \mathrm{nếu} \mathrm{b} < 0\end{array}\right.$

$\frac{\mathrm{a}}{\mathrm{b}}\sqrt{\frac{\mathrm{b}}{\mathrm{a}}} = \frac{\mathrm{a}}{\mathrm{b}}\sqrt{\frac{\mathrm{ab}}{{\mathrm{a}}^2}} = \frac{\mathrm{a}}{\mathrm{b}\left|\mathrm{a}\right|}\sqrt{\mathrm{ab}}=\left\{\begin{array}{l}\frac{1}{\mathrm{b}}\sqrt{\mathrm{ab} }\mathrm{nếu} \mathrm{b} > 0\\ -\frac{1}{\mathrm{b}}\sqrt{\mathrm{ab} }\mathrm{nếu} \mathrm{b} < 0\end{array}\right.$

$\sqrt{\frac{1}{\mathrm{b}}+\frac{1}{{\mathrm{b}}^2}} = \sqrt{\frac{\mathrm{b}+1}{{\mathrm{b}}^2}} = \frac{1}{\left|\mathrm{b}\right|}\sqrt{\mathrm{b}+1} = \left\{\begin{array}{l}\frac{1}{\mathrm{b}}\sqrt{\mathrm{b}+1 }\mathrm{nếu} \mathrm{b} > 0\\ -\frac{1}{\mathrm{b}}\sqrt{\mathrm{b}+1 }\mathrm{nếu} \mathrm{b} < 0\end{array}\right.$

$\sqrt{\frac{9{\mathrm{a}}^3}{36\mathrm{b}}} = \sqrt{\frac{{\mathrm{a}}^2.\mathrm{a}.\mathrm{b}}{\mathrm{ab}.\mathrm{b}}} = \left|\frac{\mathrm{a}}{2\mathrm{b}}\right|\sqrt{\mathrm{ab}} = \left\{\begin{array}{l}\frac{\mathrm{a}}{2\mathrm{b}}\sqrt{\mathrm{ab} }\mathrm{nếu} \mathrm{a}.\mathrm{b} > 0\\ -\frac{\mathrm{a}}{2\mathrm{b}}\sqrt{\mathrm{ab}} \mathrm{nếu} \mathrm{a}.\mathrm{b} < 0\end{array}\right.$

$3\mathrm{xy}\sqrt{\frac{2}{\mathrm{xy}}} = 3\sqrt{\frac{2{\left(\mathrm{xy}\right)}^2}{\mathrm{xy}}} = 3\sqrt{2\mathrm{xy}} \mathit{(}do\mathit{ }xy\mathit{ }\mathit{>}\mathit{ }\mathit{0}\mathit{ }nên\mathit{ }dư\mathit{ }thừa\mathit{ }số\mathit{ }xy\mathit{ }vào\mathit{ }trong\mathit{ }dấu\mathit{ }căn\mathit{)}$