Câu 49:

Khử mẫu của biểu thức lấy căn (giả thiết các biểu thức có nghĩa)

\(a b \sqrt{\frac{a}{b}} ; \frac{a}{b} \sqrt{\frac{b}{a}} ; \sqrt{\frac{1}{b}+\frac{1}{b^{2}}} ; \sqrt{\frac{9 a^{2}}{36 b}} ; 3 x z \sqrt{\frac{2}{x y}}\)

Lời giải:

+ \(a b \sqrt{\frac{a}{b}}=a b \sqrt{\frac{a \cdot b}{b \cdot b}}=a b \sqrt{\frac{a b}{b^{2}}}=\frac{a b}{|b|} \cdot \sqrt{a b}\)

\(=\left[\begin{array}{l} a \sqrt{a b} \text { khi } b>0, a>0 \\ -a \sqrt{a b} \text { khi } b<0, a \leq 0 \end{array}\right.\)

+ \(\frac{\mathrm{a}}{\mathrm{b}} \sqrt{\frac{\mathrm{b}}{\mathrm{a}}}=\frac{\mathrm{a}}{\mathrm{b}} \sqrt{\frac{\mathrm{ab}}{\mathrm{a}^{2}}}=\frac{\mathrm{a}}{\mathrm{b}|\mathrm{a}|} \sqrt{\mathrm{ab}}\)

= \(\left[\begin{array}{l} \frac{\sqrt{a b}}{b} \text { nếu } a>0 ; b>0 \\ -\frac{\sqrt{a b}}{b} \text { nếu } a<0 ; b<0 \end{array}\right.\)

+ \(\sqrt{\frac{9 a^{3}}{36 b}}=\sqrt{\frac{a^{3}}{4 b}}=\sqrt{\frac{1}{4} \cdot \frac{a^{3}}{b}}=\sqrt{\frac{1}{4}} \cdot \sqrt{\frac{a^{3}}{b}}\)

\(=\frac{1}{2} \cdot \sqrt{\frac{\mathrm{a}^{2} \cdot \mathrm{a}}{\mathrm{b}}}=\frac{1}{2} \cdot|\mathrm{a}| \cdot \sqrt{\frac{\mathrm{a}}{\mathrm{b}}}=\frac{1}{2} \cdot|\mathrm{a}| \cdot \frac{\sqrt{\mathrm{ab}}}{|\mathrm{b}|}\)

\(+\text { Nếu } \mathrm{a} \geq 0 ; \mathrm{b}>0 \Rightarrow \frac{1}{2} \cdot|\mathrm{a}| \cdot \frac{\sqrt{\mathrm{ab}}}{|\mathrm{b}|}=\frac{1}{2} \cdot \mathrm{a} \cdot \frac{\sqrt{\mathrm{ab}}}{\mathrm{b}}=\frac{\mathrm{a} \sqrt{\mathrm{ab}}}{2 \mathrm{~b}}\)

\(+\text { Nếu } \mathrm{a}<0 ; \mathrm{b}<0 \Rightarrow \frac{1}{2} \cdot|\mathrm{a}| \cdot \frac{\sqrt{\mathrm{ab}}}{|\mathrm{b}|}=\frac{1}{2} \cdot(-\mathrm{a}) \cdot \frac{\sqrt{\mathrm{ab}}}{(-\mathrm{b})}=\frac{\mathrm{a} \sqrt{\mathrm{ab}}}{2 \mathrm{~b}}\)

\(\sqrt{\frac{1}{b}+\frac{1}{b^{2}}}=\sqrt{\frac{b+1}{b^{2}}}=\frac{\sqrt{b+1}}{|b|}\)

\(=\left\{\begin{array}{l} \frac{\sqrt{b+1}}{b} \quad \text { khi } \mathrm{b}>0 \\ -\frac{\sqrt{b+1}}{\mathrm{~b}} \text { khi }-1<\mathrm{b}<0 \end{array}\right.\)

\(3 x y \sqrt{\frac{2}{x y}}=3 \sqrt{\frac{2(x y)^{2}}{x y}}=3 \sqrt{2 x y}\)

(do xy > 0 (gt) nên đưa thừa số xy vào trong căn để khử mẫu)