Bài 53 (Trang 30 SGK Toán 9, Tập 1)

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

a) $\sqrt{18{(\sqrt{2}-\sqrt{3})}^2};$

b) $\mathrm{ab}\sqrt{1+\frac{1}{{\mathrm{a}}^2{\mathrm{b}}^2}}$;

c) $\sqrt{\frac{\mathrm{a}}{{\mathrm{b}}^3}+\frac{\mathrm{a}}{{\mathrm{b}}^4}}$;

 d) $\frac{\mathrm{a}+\sqrt{\mathrm{ab}}}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}$.

Hướng dẫn giải:

a) $\sqrt{18{(\sqrt{2}-\sqrt{3})}^2} = \left|\sqrt{2}-\sqrt{3}\right|\sqrt{2.9}$

  $= (\sqrt{3}-\sqrt{2}).\sqrt{2.3^2} = 3(\sqrt{3}-\sqrt{2})\sqrt{2}$

b) $\mathrm{ab}\sqrt{1+\frac{1}{{\mathrm{a}}^2{\mathrm{b}}^2}} = \mathrm{ab}\sqrt{\frac{1+{\mathrm{a}}^2{\mathrm{b}}^2}{{\mathrm{a}}^2{\mathrm{b}}^2}} = \frac{\mathrm{ab}}{\left|\mathrm{ab}\right|}\sqrt{1+{\mathrm{a}}^2{\mathrm{b}}^2}$

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

c) $\sqrt{\frac{\mathrm{a}}{{\mathrm{b}}^3}+\frac{\mathrm{a}}{{\mathrm{b}}^4}} = \sqrt{\frac{\mathrm{a}+\mathrm{ab}}{{\mathrm{b}}^4}} = \frac{1}{{\mathrm{b}}^2}\sqrt{\mathrm{a}+\mathrm{ab}};$

d) $\frac{\mathrm{a}+\sqrt{\mathrm{ab}}}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}} = \frac{(\mathrm{a}+\sqrt{\mathrm{ab}})(\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}})}{(\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}})(\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}})} = \frac{\mathrm{a}\sqrt{\mathrm{a}}-\mathrm{a}\sqrt{\mathrm{b}}+\mathrm{a}\sqrt{\mathrm{b}}-\mathrm{b}\sqrt{\mathrm{a}}}{\mathrm{a}-\mathrm{b}}$

   =$\frac{a\sqrt{a}-b\sqrt{a}}{a-b} = \frac{(a-b)\sqrt{a}}{a-b} = \sqrt{a};$

Cách khác: $\frac{\mathrm{a}+\sqrt{\mathrm{ab}}}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}} = \frac{{\left(\sqrt{\mathrm{a}}\right)}^2+\sqrt{\mathrm{a}}.\sqrt{\mathrm{b}}}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}} = \frac{\sqrt{\mathrm{a}}(\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}})}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}} = \sqrt{\mathrm{a}}.$