登录
6. 先化简，再求值：$\dfrac{3}{x}\sqrt{x^{3} - 2x^{2}} - (x - 3)\sqrt{\dfrac{4}{x - 3}}$，其中$x = 21$.
答案
6. 原式$ =3\sqrt{x - 2}-2\sqrt{x - 3} $，当$ x = 21 $时，原式$ =3\sqrt{19}-6\sqrt{2} $。
解析
解：原式$=\dfrac{3}{x}\sqrt{x^{2}(x - 2)} - (x - 3)\sqrt{\dfrac{4(x - 3)}{(x - 3)^{2}}}$
$=\dfrac{3}{x}·|x|\sqrt{x - 2} - (x - 3)·\dfrac{2\sqrt{x - 3}}{|x - 3|}$
由二次根式有意义的条件得：$x - 3 ＞ 0$，即$x ＞ 3$，则$|x| = x$，$|x - 3| = x - 3$
所以原式$=\dfrac{3}{x}· x\sqrt{x - 2} - (x - 3)·\dfrac{2\sqrt{x - 3}}{x - 3}=3\sqrt{x - 2}-2\sqrt{x - 3}$
当$x = 21$时，原式$=3\sqrt{21 - 2}-2\sqrt{21 - 3}=3\sqrt{19}-2\sqrt{18}=3\sqrt{19}-6\sqrt{2}$
$=\dfrac{3}{x}·|x|\sqrt{x - 2} - (x - 3)·\dfrac{2\sqrt{x - 3}}{|x - 3|}$
由二次根式有意义的条件得：$x - 3 ＞ 0$，即$x ＞ 3$，则$|x| = x$，$|x - 3| = x - 3$
所以原式$=\dfrac{3}{x}· x\sqrt{x - 2} - (x - 3)·\dfrac{2\sqrt{x - 3}}{x - 3}=3\sqrt{x - 2}-2\sqrt{x - 3}$
当$x = 21$时，原式$=3\sqrt{21 - 2}-2\sqrt{21 - 3}=3\sqrt{19}-2\sqrt{18}=3\sqrt{19}-6\sqrt{2}$
7. 已知$x = \sqrt{3} - \sqrt{2}$，$y = \sqrt{3} + \sqrt{2}$，求$x^{3}y + xy^{3}$的值.
答案
7. 10
解析
$x^{3}y + xy^{3} = xy(x^{2} + y^{2})$
$x + y = (\sqrt{3} - \sqrt{2}) + (\sqrt{3} + \sqrt{2}) = 2\sqrt{3}$
$xy = (\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2}) = (\sqrt{3})^{2} - (\sqrt{2})^{2} = 3 - 2 = 1$
$x^{2} + y^{2} = (x + y)^{2} - 2xy = (2\sqrt{3})^{2} - 2×1 = 12 - 2 = 10$
$x^{3}y + xy^{3} = 1×10 = 10$
$x + y = (\sqrt{3} - \sqrt{2}) + (\sqrt{3} + \sqrt{2}) = 2\sqrt{3}$
$xy = (\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2}) = (\sqrt{3})^{2} - (\sqrt{2})^{2} = 3 - 2 = 1$
$x^{2} + y^{2} = (x + y)^{2} - 2xy = (2\sqrt{3})^{2} - 2×1 = 12 - 2 = 10$
$x^{3}y + xy^{3} = 1×10 = 10$
