10. (2025·南宁期末)已知$x$是整数,$\sqrt{3} · \sqrt{\dfrac{6}{x}}$是整数,则$x$的最小值是 (
A.2
B.3
C.4
D.18
A
)A.2
B.3
C.4
D.18
答案：10.A
11. (2025·东营期末)计算:$(1+\sqrt{2})^{2024}(1-\sqrt{2})^{2025}=$
$1-\sqrt{2}$
.答案：11.$1-\sqrt{2}$
12. 计算$\sqrt{-2a} · \sqrt{-8a}(a＜0)$的结果是
$-4a$
.答案：12.$-4a$
13. 观察下列各等式:$\sqrt{1} × \sqrt{3}=\sqrt{2^{2}-1}$,$\sqrt{2} × \sqrt{4}=\sqrt{3^{2}-1}$,$\sqrt{3} × \sqrt{5}=\sqrt{4^{2}-1}$,$\sqrt{4} × \sqrt{6}=$$\sqrt{5^{2}-1}$,$······$.请用含$n(n≥ 1$且$n$为整数)的等式表示你所观察到的规律:
$\sqrt{n}×\sqrt{n+2}=\sqrt{(n+1)^{2}-1}$
.答案：13.$\sqrt{n}×\sqrt{n+2}=\sqrt{(n+1)^{2}-1}$
14. 计算:
(1)$\sqrt{3} × \sqrt{12}+|-2|-(π-3)^{0}$; (2)$\dfrac{3}{2}\sqrt{20} × (-\sqrt{15}) × (-\dfrac{1}{3}\sqrt{27})$;
(3)$-\sqrt{18} × \sqrt{(-54) × (-\dfrac{1}{27})}$; (4)$\sqrt{2-\sqrt{3}} × \sqrt{2+\sqrt{3}}$.
(1)$\sqrt{3} × \sqrt{12}+|-2|-(π-3)^{0}$; (2)$\dfrac{3}{2}\sqrt{20} × (-\sqrt{15}) × (-\dfrac{1}{3}\sqrt{27})$;
(3)$-\sqrt{18} × \sqrt{(-54) × (-\dfrac{1}{27})}$; (4)$\sqrt{2-\sqrt{3}} × \sqrt{2+\sqrt{3}}$.
答案：14.解:(1)原式$=6+2-1=7$.
(2)原式$=\frac{3}{2}×2\sqrt{5}×\sqrt{15}×\frac{1}{3}×3\sqrt{3}=45$.
(3)原式$=-3\sqrt{2}×\sqrt{2}=-6$.
(4)原式$=\sqrt{(2-\sqrt{3})(2+\sqrt{3})}=\sqrt{4-3}=1$.
(2)原式$=\frac{3}{2}×2\sqrt{5}×\sqrt{15}×\frac{1}{3}×3\sqrt{3}=45$.
(3)原式$=-3\sqrt{2}×\sqrt{2}=-6$.
(4)原式$=\sqrt{(2-\sqrt{3})(2+\sqrt{3})}=\sqrt{4-3}=1$.
15. 你能找出规律吗?
(1)计算:$\sqrt{9} × \sqrt{16}=$
(2)请按找到的规律计算:
①$\sqrt{5} × \sqrt{125}$;②$\sqrt{1\dfrac{2}{3}} × \sqrt{9\dfrac{3}{5}}$.
(3)已知$a=\sqrt{2}$,$b=\sqrt{10}$,用含$a,b$的式子表示$\sqrt{180}$.
(1)计算:$\sqrt{9} × \sqrt{16}=$
12
,$\sqrt{9 × 16}=$12
;$\sqrt{25} × \sqrt{36}=$30
,$\sqrt{25 × 36}=$30
.(2)请按找到的规律计算:
①$\sqrt{5} × \sqrt{125}$;②$\sqrt{1\dfrac{2}{3}} × \sqrt{9\dfrac{3}{5}}$.
(3)已知$a=\sqrt{2}$,$b=\sqrt{10}$,用含$a,b$的式子表示$\sqrt{180}$.
答案：15.(1)12 12 30 30
(2)解:①$\sqrt{5}×\sqrt{125}=\sqrt{5×125}=\sqrt{625}=25$.
②$\sqrt{1\frac{2}{3}}×\sqrt{9\frac{3}{5}}=\sqrt{\frac{5}{3}×\frac{48}{5}}=\sqrt{16}=4$.
(3)解:当$a=\sqrt{2},b=\sqrt{10}$时,
$\sqrt{180}=\sqrt{9×2×10}=\sqrt{9}×\sqrt{2}×\sqrt{10}=3ab$.
(2)解:①$\sqrt{5}×\sqrt{125}=\sqrt{5×125}=\sqrt{625}=25$.
②$\sqrt{1\frac{2}{3}}×\sqrt{9\frac{3}{5}}=\sqrt{\frac{5}{3}×\frac{48}{5}}=\sqrt{16}=4$.
(3)解:当$a=\sqrt{2},b=\sqrt{10}$时,
$\sqrt{180}=\sqrt{9×2×10}=\sqrt{9}×\sqrt{2}×\sqrt{10}=3ab$.
16. 比较下列各组值的大小:(在横线上填“＞”“＜”或“=”)
$4+3\_\_\_\_\_\_2 × \sqrt{4} × \sqrt{3}$;$3+\dfrac{1}{2}\_\_\_\_\_\_2 × \sqrt{3} × \sqrt{\dfrac{1}{2}}$;$5+5\_\_\_\_\_\_2 × \sqrt{5} × \sqrt{5}$;$······$.
通过观察归纳,写出能反映这种规律的一般结论,并说明你所写式子的正确性.
$4+3\_\_\_\_\_\_2 × \sqrt{4} × \sqrt{3}$;$3+\dfrac{1}{2}\_\_\_\_\_\_2 × \sqrt{3} × \sqrt{\dfrac{1}{2}}$;$5+5\_\_\_\_\_\_2 × \sqrt{5} × \sqrt{5}$;$······$.
通过观察归纳,写出能反映这种规律的一般结论,并说明你所写式子的正确性.
答案：16.$＞$ $＞$ $=$
解:结论为$a+b≥ 2\sqrt{ab}(a≥ 0,b≥ 0)$.
证明:$\because a+b-2\sqrt{ab}=(\sqrt{a}-\sqrt{b})^{2}≥ 0,\therefore a+b≥ 2\sqrt{ab}$.
解:结论为$a+b≥ 2\sqrt{ab}(a≥ 0,b≥ 0)$.
证明:$\because a+b-2\sqrt{ab}=(\sqrt{a}-\sqrt{b})^{2}≥ 0,\therefore a+b≥ 2\sqrt{ab}$.