1. 下列运算正确的是 (
C
)
A.$\sqrt{2}+\sqrt{3}=\sqrt{5}$
B.$\sqrt{(-5)^{2}}=-5$
C.$(3-\sqrt{2})^{2}=11-6\sqrt{2}$
D.$6÷ \frac{2}{\sqrt{3}}× \sqrt{3}=3$
答案：1.C
2. (2025·河北)计算:$(\sqrt{10}+\sqrt{6})(\sqrt{10}-\sqrt{6})=$ (
B
)
A.2
B.4
C.6
D.8
答案：2.B
3. 计算:(1)$\sqrt{3}×(\sqrt{12}+2\sqrt{15})=$
$6+6\sqrt{5}$
; (2)$(\sqrt{18}-\sqrt{8})×\sqrt{2}=$
2
;
(3)$(2-\sqrt{3})^{2}=$
$7-4\sqrt{3}$
; (4)$(2+\sqrt{3})(2-\sqrt{3})=$
1
.
答案：3.(1)$6+6\sqrt{5}$ (2)2 (3)$7-4\sqrt{3}$ (4)1
4. 已知 $a=3+2\sqrt{2},b=3-2\sqrt{2}$,则 $a^{2}b-ab^{2}=$
$4\sqrt{2}$
.
答案：4.$4\sqrt{2}$
5. (2025·天津)计算$(\sqrt{61}+1)(\sqrt{61}-1)$的结果为
60
.
答案：5.60
6. 计算:
(1)$\sqrt{8}+\sqrt{2}×(1+\sqrt{2})$;
(2)$2\sqrt{3}×(\sqrt{72}-2\sqrt{50})$;
(3)$(\sqrt{3}+1)×(\sqrt{3}-2)$;
(4)$(3\sqrt{2}-2\sqrt{3})×(2\sqrt{2}+3\sqrt{3})$.
答案：6.解:(1)原式$=2\sqrt{2}+\sqrt{2}+2=2+3\sqrt{2}$.
(2)原式$=2\sqrt{3}×(6\sqrt{2}-10\sqrt{2})=2\sqrt{3}×(-4\sqrt{2})=-8\sqrt{6}$.
(3)原式$=3-2\sqrt{3}+\sqrt{3}-2=1-\sqrt{3}$.
(4)原式$=12+9\sqrt{6}-4\sqrt{6}-18=5\sqrt{6}-6$.
7. (2025·邢台期末)已知 $x=\sqrt{5}-2,y=\sqrt{5}+2$,则 $(\frac{x}{y}-1)· \frac{y}{x^{2}-y^{2}}$的值为 (
D
)
A.$\frac{1}{4}$
B.$\frac{\sqrt{5}}{5}$
C.$\frac{2\sqrt{5}}{5}$
D.$\frac{\sqrt{5}}{10}$
答案：7.D
8. (2025·佛山一模)若 $3-\sqrt{2}$的整数部分为$a$,小数部分为$b$,则$(2+\sqrt{2}a)· b=$ (
C
)
A.-1
B.$-\sqrt{2}$
C.2
D.0
答案：8.C
9. 计算下列各式:
(1)$(1+\sqrt{2}+\sqrt{3})×(1+\sqrt{2}-\sqrt{3})$;
(2)$(\sqrt{2}+\sqrt{3})^{2}-(\sqrt{2}-\sqrt{3})^{2}$;
(3)$(\sqrt{5}-2)^{2}-(\sqrt{13}-2)(\sqrt{13}+2)$;
(4)$[(\sqrt{a}-\sqrt{b})^{2}+4\sqrt{ab}]÷(\sqrt{a}+\sqrt{b})$.
答案：9.解:(1)原式$=(1+\sqrt{2})^{2}-(\sqrt{3})^{2}=3+2\sqrt{2}-3=2\sqrt{2}$.
(2)原式$=(5+2\sqrt{6})-(5-2\sqrt{6})=4\sqrt{6}$.
(3)原式$=5-4\sqrt{5}+4-(13-4)=9-4\sqrt{5}-9=-4\sqrt{5}$.
(4)原式$=(\sqrt{a}+\sqrt{b})^{2}÷(\sqrt{a}+\sqrt{b})=\sqrt{a}+\sqrt{b}$.
