Bài 75 (Trang 41 SGK Toán 9, Tập 1):

Chứng minh các đẳng thức sau:

$\mathrm{a}) \left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2} - \frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}} = -1,5;$

$\mathrm{b}) \left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}= -2;$

$\mathrm{c}) \frac{\mathrm{a}\sqrt{\mathrm{b}}+\mathrm{b}\sqrt{\mathrm{a}}}{\sqrt{\mathrm{ab}}}:\frac{1}{\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}}} = \mathrm{a} - \mathrm{b} \mathrm{với} \mathrm{a}, \mathrm{b} > 0 \mathrm{và} \mathrm{a}\ne \mathrm{b};$

$\mathrm{d}) \left(1 + \frac{\mathrm{a}+\sqrt{\mathrm{a}}}{\sqrt{\mathrm{a}}+1}\right)\left(1 - \frac{\mathrm{a}-\sqrt{\mathrm{a}}}{\sqrt{\mathrm{a}}-1}\right) = 1 - \mathrm{a} với a \ge 0 và a \ne 1.$

Hướng dẫn giải:

a)

$$\begin{gathered} \mathrm{VT} = \left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2} - \frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}} \\ = \left(\frac{\sqrt{2^2.3}-\sqrt{6}}{\sqrt{4.2}-2} -\frac{\sqrt{36.6}}{3}\right)\frac{1}{\sqrt{6}} \\ = \left(\frac{\sqrt{2}.\sqrt{6}-\sqrt{6}}{2\left(\sqrt{2}-1\right)} - 2\sqrt{6}\right)\frac{1}{\sqrt{6}} \\ =\left(\frac{\sqrt{6}}{2} - 2\sqrt{6}\right)\frac{1}{\sqrt{6}} \\ =\frac{-3\sqrt{6}}{2}.\frac{1}{\sqrt{6}} = 1,5 = \mathrm{VP} \end{gathered}$$

b)

$$\begin{gathered} \mathrm{VT} = \left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}} + \frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}} = -2 \\ = \left[\frac{\sqrt{7}.\sqrt{2}-\sqrt{7}}{1 -\sqrt{2}}+\frac{\sqrt{5}.\sqrt{3}-\sqrt{5}}{1-\sqrt{3}}\right]:\frac{1}{\sqrt{7}-\sqrt{5}} \\ =\left[\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right].\left(\sqrt{7}-\sqrt{5}\right) \\ =\left(-\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right) \\ =-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right) \\ =-\left(7-5\right) = -2 = \mathrm{VP} \end{gathered}$$

c)

$$\begin{gathered} \mathrm{VT} = \frac{\mathrm{a}\sqrt{\mathrm{b}}+\mathrm{b}\sqrt{\mathrm{a}}}{\sqrt{\mathrm{ab}}}:\frac{1}{\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}}} \\ = \frac{\sqrt{a^2}.\sqrt{b}+\sqrt{b^2.\sqrt{a}}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}} \\ =\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right) \\ =\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right) = a - b = VP \end{gathered}$$

d)

$$\begin{gathered} \mathrm{VT} = \left(1 + \frac{\mathrm{a}+\sqrt{\mathrm{a}}}{\sqrt{\mathrm{a}}+1}\right)\left(1 - \frac{\mathrm{a}-\sqrt{\mathrm{a}}}{\sqrt{\mathrm{a}}-1}\right) \\ = \left(1 + \frac{\sqrt{{\mathrm{a}}^2}+\sqrt{\mathrm{a}}}{\sqrt{\mathrm{a}}+1}\right)\left(1 - \frac{\sqrt{{\mathrm{a}}^2}-\sqrt{\mathrm{a}}}{\sqrt{\mathrm{a}}-1}\right) \\ = \left[1 + \frac{\sqrt{\mathrm{a}}\left(\sqrt{\mathrm{a}}+1\right)}{\sqrt{\mathrm{a}}+1}\right]\left[1 - \frac{\sqrt{\mathrm{a}}\left(\sqrt{\mathrm{a}}-1\right)}{\sqrt{\mathrm{a}}-1}\right] \\ = \left(1 + \sqrt{\mathrm{a}}\right)\left(1 - \sqrt{\mathrm{a}}\right) = 1 - \left(\sqrt{{\mathrm{a}}^2}\right) = 1-1 =\mathrm{VP} \end{gathered}$$