Câu 49: Trang 29 - sgk toán 9 tập 1
Khử mẫu của biểu thức lấy căn :
$ab\sqrt{\frac{a}{b}}$ ; $\frac{a}{b}\sqrt{\frac{b}{a}}$ ; $\sqrt{\frac{1}{b}+\frac{1}{b^{2}}}$ ; $\sqrt{\frac{9a^{3}}{36b}}$ ; $3xy\sqrt{\frac{2}{xy}}$
Bài Làm:
Ta có :
- $ab\sqrt{\frac{a}{b}}=ab\sqrt{\frac{ab}{b^{2}}}=\frac{ab\sqrt{ab}}{\left | b \right |}=\left\{\begin{matrix}a\sqrt{ab}(a\geq 0,b>0) & \\ -a\sqrt{ab}(a\leq 0,b<0) & \end{matrix}\right.$
- $\frac{a}{b}\sqrt{\frac{b}{a}}=\frac{a}{b}\sqrt{\frac{ab}{b^{2}}}=\frac{a\sqrt{ab}}{b\left | a \right|}=\left\{\begin{matrix}\frac{\sqrt{ab}}{b}(a,b>0) & \\ \frac{-\sqrt{ab}}{b}(a,b<0) & \end{matrix}\right.$
- $\sqrt{\frac{1}{b}+\frac{1}{b^{2}}}=\sqrt{\frac{b+1}{b^{2}}}=\frac{\sqrt{b+1}}{\left | b \right |}=\left\{\begin{matrix}\frac{\sqrt{b+1}}{b}(b>0) & \\ \frac{-\sqrt{b+1}}{b}(-1\leq b<0) & \end{matrix}\right.$
- $\sqrt{\frac{9a^{3}}{36b}}=\sqrt{\frac{9a^{3}b}{36b^{2}}}=\frac{3\left | a \right |\sqrt{ab}}{6\left | b \right |}=\frac{a\sqrt{ab}}{2b}$
- $3xy\sqrt{\frac{2}{xy}}=3xy\sqrt{\frac{2xy}{(xy)^{2}}}=\frac{3xy\sqrt{2xy}}{\left | xy \right |}=3\sqrt{2xy}$