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7. 化简:$\sqrt{x^{4}(x^{2}+1)}=$.
答案
7. $x^{2}\sqrt{x^{2}+1}$
8. 化简:$(\sqrt{a + b}+\sqrt{a - b})^{2}+(\sqrt{a + b}-\sqrt{a - b})^{2}=$.
答案
8. $4a$
解析
$\begin{aligned}&(\sqrt{a + b}+\sqrt{a - b})^{2}+(\sqrt{a + b}-\sqrt{a - b})^{2}\\=&(a + b + 2\sqrt{(a + b)(a - b)} + a - b) + (a + b - 2\sqrt{(a + b)(a - b)} + a - b)\\=&(2a + 2\sqrt{a^2 - b^2}) + (2a - 2\sqrt{a^2 - b^2})\\=&4a\end{aligned}$
1. 已知$\sqrt{18n}$是整数, 则正整数$n$的最小值为 (
A.$4$
B.$3$
C.$2$
D.$1$
C
)A.$4$
B.$3$
C.$2$
D.$1$
答案
1. C
解析
$\sqrt{18n} = \sqrt{9 × 2n} = 3\sqrt{2n}$，要使$\sqrt{18n}$是整数，则$\sqrt{2n}$必须是整数，即$2n$是完全平方数。正整数$n$最小为$2$时，$2n = 4$，$\sqrt{4} = 2$，此时$\sqrt{18n} = 3×2 = 6$是整数。C
2. 计算$(\sqrt{3}+\sqrt{2})^{2}(\sqrt{3}-\sqrt{2})^{2}$的结果是 (
A.$-1$
B.$1$
C.$\sqrt{3}-\sqrt{2}$
D.$\sqrt{3}+\sqrt{2}$
B
)A.$-1$
B.$1$
C.$\sqrt{3}-\sqrt{2}$
D.$\sqrt{3}+\sqrt{2}$
答案
2. B
解析
$\begin{aligned}&(\sqrt{3}+\sqrt{2})^{2}(\sqrt{3}-\sqrt{2})^{2}\\=&[(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})]^{2}\\=&[(\sqrt{3})^{2}-(\sqrt{2})^{2}]^{2}\\=&(3 - 2)^{2}\\=&1^{2}\\=&1\end{aligned}$
B
B
3. 若$\sqrt{(x - 2)(x - 3)}=\sqrt{x - 2}·\sqrt{3 - x}$成立, 则$x$的取值范围是 (
A.$x≥2$
B.$x≤3$
C.$2≤ x≤3$
D.$2 ＜ x ＜ 3$
C
)A.$x≥2$
B.$x≤3$
C.$2≤ x≤3$
D.$2 ＜ x ＜ 3$
答案
3. C
解析
要使$\sqrt{(x - 2)(x - 3)}=\sqrt{x - 2}·\sqrt{3 - x}$成立，需满足：
1. 根号下的数非负，即$x - 2 ≥ 0$，解得$x ≥ 2$；
2. 根号下的数非负，即$3 - x ≥ 0$，解得$x ≤ 3$；
3. 等式左边$\sqrt{(x - 2)(x - 3)} = \sqrt{(x - 2)(-(3 - x))} = \sqrt{-(x - 2)(3 - x)}$，右边$\sqrt{x - 2}·\sqrt{3 - x} = \sqrt{(x - 2)(3 - x)}$，要使等式成立，需$(x - 2)(3 - x) = 0$，即$x = 2$或$x = 3$，结合前两个条件，$2 ≤ x ≤ 3$。
C
1. 根号下的数非负，即$x - 2 ≥ 0$，解得$x ≥ 2$；
2. 根号下的数非负，即$3 - x ≥ 0$，解得$x ≤ 3$；
3. 等式左边$\sqrt{(x - 2)(x - 3)} = \sqrt{(x - 2)(-(3 - x))} = \sqrt{-(x - 2)(3 - x)}$，右边$\sqrt{x - 2}·\sqrt{3 - x} = \sqrt{(x - 2)(3 - x)}$，要使等式成立，需$(x - 2)(3 - x) = 0$，即$x = 2$或$x = 3$，结合前两个条件，$2 ≤ x ≤ 3$。
C
4. 设矩形的长和宽分别为$a,b$, 根据下列条件求面积$S$.
(1)$a=\sqrt{12},b=\sqrt{8}$;
(2)$a = 3\sqrt{24},b=\frac{1}{2}\sqrt{48}$.
(1)$a=\sqrt{12},b=\sqrt{8}$;
(2)$a = 3\sqrt{24},b=\frac{1}{2}\sqrt{48}$.
答案
(1) $ S = a · b = \sqrt{12} · \sqrt{8} = \sqrt{12 × 8} = \sqrt{96} = 4\sqrt{6} $
(2) $ S = a · b = 3\sqrt{24} · \frac{1}{2}\sqrt{48} = \frac{3}{2} × \sqrt{24 × 48} = \frac{3}{2} × \sqrt{1152} = \frac{3}{2} × 24\sqrt{2} = 36\sqrt{2} $
(2) $ S = a · b = 3\sqrt{24} · \frac{1}{2}\sqrt{48} = \frac{3}{2} × \sqrt{24 × 48} = \frac{3}{2} × \sqrt{1152} = \frac{3}{2} × 24\sqrt{2} = 36\sqrt{2} $
解析
(1) $ S = a · b = \sqrt{12} · \sqrt{8} = \sqrt{12 × 8} = \sqrt{96} = 4\sqrt{6} $
(2) $ S = a · b = 3\sqrt{24} · \frac{1}{2}\sqrt{48} = \frac{3}{2} × \sqrt{24 × 48} = \frac{3}{2} × \sqrt{1152} = \frac{3}{2} × 24\sqrt{2} = 36\sqrt{2} $
(2) $ S = a · b = 3\sqrt{24} · \frac{1}{2}\sqrt{48} = \frac{3}{2} × \sqrt{24 × 48} = \frac{3}{2} × \sqrt{1152} = \frac{3}{2} × 24\sqrt{2} = 36\sqrt{2} $
5. 比较下列各组中两个数的大小.
(1)$2\sqrt{7}$和$4\sqrt{2}$;
(2)$2\sqrt{3}$和$3\sqrt{2}$.
(1)$2\sqrt{7}$和$4\sqrt{2}$;
(2)$2\sqrt{3}$和$3\sqrt{2}$.
答案
5. (1) $2\sqrt{7}＜4\sqrt{2}$ (2) $2\sqrt{3}＜3\sqrt{2}$
解析
(1) 因为 $(2\sqrt{7})^2 = 4 × 7 = 28$，$(4\sqrt{2})^2 = 16 × 2 = 32$，且 $28 ＜ 32$，所以 $2\sqrt{7} ＜ 4\sqrt{2}$；
(2) 因为 $(2\sqrt{3})^2 = 4 × 3 = 12$，$(3\sqrt{2})^2 = 9 × 2 = 18$，且 $12 ＜ 18$，所以 $2\sqrt{3} ＜ 3\sqrt{2}$。
(2) 因为 $(2\sqrt{3})^2 = 4 × 3 = 12$，$(3\sqrt{2})^2 = 9 × 2 = 18$，且 $12 ＜ 18$，所以 $2\sqrt{3} ＜ 3\sqrt{2}$。
6. 先化简, 再求值:$6x^{2}+2xy - 8y^{2}-2(3xy - 4y^{2}+3x^{2})$, 其中$x=\sqrt{2},y=\sqrt{6}$.
答案
6. 解:原式$=6x^{2}+2xy - 8y^{2}-6xy + 8y^{2}-6x^{2}=(6x^{2}-6x^{2})+(2xy - 6xy)+(-8y^{2}+8y^{2})=-4xy$.当$x=\sqrt{2},y=\sqrt{6}$时,$-4xy=-4×\sqrt{2}×\sqrt{6}=-8\sqrt{3}$.
