Bài 64 (Trang 33 SGK Toán 9, Tập 1):

Chứng minh các đẳng thức sau:

$\mathrm{a}) \left(\frac{1-\mathrm{a}\sqrt{\mathrm{a}}}{1-\sqrt{\mathrm{a}}}+\sqrt{\mathrm{a}}\right){\left(\frac{1-\sqrt{\mathrm{a}}}{1-\mathrm{a}}\right)}^2 = 1 với\mathit{ }a\mathit{ }\mathit{>}\mathit{ }\mathit{0}\mathit{ }và\mathit{ }a\mathit{ }\mathit{\ne }\mathit{1}\mathit{;}$

$\mathrm{b}) \frac{\mathrm{a} + \mathrm{b}}{{\mathrm{b}}^2}. \sqrt{\frac{{\mathrm{a}}^2{\mathrm{b}}^4}{{\mathrm{a}}^{2 }+ 2\mathrm{ab} + {\mathrm{b}}^2}} =\left|\mathrm{a}\right| với\mathit{ }a\mathit{ }\mathit{+}\mathit{ }b\mathit{ }\mathit{>}\mathit{ }\mathit{0}\mathit{ }và\mathit{ }b\mathit{\ne }\mathit{0}\mathit{.}$

Hướng dẫn giải:

a) Thực hiện vế trái:

$$\begin{gathered} \mathrm{VT} = \left(\frac{1-\mathrm{a}\sqrt{\mathrm{a}}}{1-\sqrt{\mathrm{a}}} +\sqrt{\mathrm{a}}\right){\left(\frac{1-\sqrt{\mathrm{a}}}{1-\mathrm{a}}\right)}^2= \frac{1-\mathrm{a}\sqrt{\mathrm{a}}+\sqrt{\mathrm{a}}-\mathrm{a}}{1-\sqrt{\mathrm{a}}}.\frac{{\left(1-\sqrt{\mathrm{a}}\right)}^2}{{\left(1-\mathrm{a}\right)}^2} \\ = \left(1-\mathrm{a}\sqrt{\mathrm{a}}+ \sqrt{\mathrm{a}} -\mathrm{a}\right)\frac{1-\sqrt{\mathrm{a}}}{{\left(1-\mathrm{a}\right)}^2} = \frac{1-\mathrm{a}\sqrt{\mathrm{a}}+\sqrt{\mathrm{a}}-\mathrm{a}-\sqrt{\mathrm{a}}+\mathrm{a}.\sqrt{{\mathrm{a}}^2}-{\left(\sqrt{\mathrm{a}}\right)}^2+\mathrm{a}\sqrt{\mathrm{a}}}{{\left(1-\mathrm{a}\right)}^2} \\ = \frac{{\mathrm{a}}^2-2\mathrm{a}+1}{{\left(1-\mathrm{a}\right)}^2}=\frac{{\left(\mathrm{a}-1\right)}^2}{{\left(1-\mathrm{a}\right)}^2} = {\left(\frac{\mathrm{a}-1}{1-\mathrm{a}}\right)}^2= \left(-1\right) =1= \mathrm{VP} (đẳng thức được chứng minh) \end{gathered}$$

b) Thực hiện vế trái:

$$\begin{gathered} \mathrm{VT} = \frac{\mathrm{a} + \mathrm{b}}{{\mathrm{b}}^2}. \sqrt{\frac{{\mathrm{a}}^2{\mathrm{b}}^4}{{\mathrm{a}}^2 + 2\mathrm{ab} + {\mathrm{b}}^2}}= \frac{\mathrm{a} + \mathrm{b}}{{\mathrm{b}}^2}.\sqrt{\frac{{\left({\mathrm{ab}}^2\right)}^2}{{\left(\mathrm{a} + \mathrm{b}\right)}^2}} \\ = \frac{\mathrm{a}+\mathrm{b}}{{\mathrm{b}}^2}.\frac{\left|{\mathrm{ab}}^2\right|}{\left|\mathrm{a}+\mathrm{b}\right|} = \frac{\mathrm{a}+\mathrm{b}}{{\mathrm{b}}^2}.\frac{{\mathrm{b}}^2\left|\mathrm{a}\right|}{\mathrm{a}+\mathrm{b}} =\left|\mathrm{a}\right| = \mathrm{VP} \mathit{(}đẳng\mathit{ }thức\mathit{ }được\mathit{ }chứng\mathit{ }minh\mathit{)} \end{gathered}$$