Bài 72 (Trang 40 SGK Toán 9, Tập 1):

Phân tích thành nhân tử (với các số x, y, a, b không âm và a ≥ b)

$\mathrm{a}) \mathrm{xy} - \mathrm{y}\sqrt{\mathrm{x}} +\hspace{0.17em}\sqrt{\mathrm{x}} - 1;$

$\mathrm{b}) \sqrt{\mathrm{ax}} - \sqrt{\mathrm{by}} + \sqrt{\mathrm{bx}} - \sqrt{\mathrm{ay}};$

$\mathrm{c}) \sqrt{\mathrm{a} + \mathrm{b}} + \sqrt{{\mathrm{a}}^2 - {\mathrm{b}}^2};$

$\mathrm{d}) 12 - \sqrt{\mathrm{x}} - \mathrm{x}.$

Hướng dẫn giải:

$$\begin{gathered} \mathrm{a}) \mathrm{xy} - \mathrm{y}\sqrt{\mathrm{x}} + \sqrt{\mathrm{x}} - 1 = {\left(\sqrt{\mathrm{x}}\right)}^2.\mathrm{y} - \mathrm{y}\sqrt{\mathrm{x}} + \sqrt{\mathrm{x}} + 1 \\ = \mathrm{y}\sqrt{\mathrm{x}} - \mathrm{y}\sqrt{\mathrm{x}} + \sqrt{\mathrm{x}} - 1 = \mathrm{y}\sqrt{\mathrm{x}}\left(\sqrt{\mathrm{x}} - 1\right) + \sqrt{\mathrm{x}} - 1 \\ = \left(\sqrt{\mathrm{x}} - 1\right) \mathrm{với} \mathrm{x} \ge 0. \end{gathered}$$

$$\begin{gathered} \mathrm{b}) \sqrt{\mathrm{ax}} - \sqrt{\mathrm{by}} + \sqrt{\mathrm{bx}} - \sqrt{\mathrm{ay}} = \sqrt{\mathrm{ax}} - \sqrt{\mathrm{bx}} + \sqrt{\mathrm{ay}} - \sqrt{\mathrm{by}} \\ = \sqrt{\mathrm{x}}\left(\sqrt{\mathrm{a}} + \sqrt{\mathrm{b}}\right) - \sqrt{\mathrm{y}}\left(\sqrt{\mathrm{a}} +\hspace{0.17em}\sqrt{\mathrm{b}}\right) \\ = \left(\sqrt{\mathrm{a}} + \sqrt{\mathrm{b}}\right)\left(\sqrt{\mathrm{x}} - \sqrt{\mathrm{y}}\right) \\ \mathit{(}Với\mathit{ }x\mathit{,}\mathit{ }y\mathit{,}\mathit{ }a\mathit{,}\mathit{ }b\mathit{ }đều\mathit{ }không\mathit{ }âm\mathit{)} \end{gathered}$$

$$\begin{gathered} \mathrm{c}) \sqrt{\mathrm{a} + \mathrm{b}} - \sqrt{{\mathrm{a}}^2 - {\mathrm{b}}^2} = \sqrt{\mathrm{a} + \mathrm{b}} + \sqrt{\left(\mathrm{a} - \mathrm{b}\right)\left(\mathrm{a} + \mathrm{b}\right)} \\ = \sqrt{\mathrm{a} + \mathrm{b}} + \sqrt{\mathrm{a} - \mathrm{b}}.\sqrt{\mathrm{a} + \mathrm{b}} \\ = \sqrt{\mathrm{a} + \mathrm{b}}\left(1 + \sqrt{\mathrm{a} - \mathrm{b}}\right) \\ \mathit{(}Với\mathit{ }a\mathit{ }\mathit{+}\mathit{ }b\mathit{,}\mathit{ }a\mathit{ }\mathit{-}\mathit{ }b\mathit{ }không\mathit{ }âm\mathit{)} \end{gathered}$$