1. 已知$a=2-\sqrt{3}$,则$\sqrt{a^{2}-2a+1}$的值为 (
A.$1-\sqrt{3}$
B.$\sqrt{3}-1$
C.$3-\sqrt{3}$
D.$\sqrt{3}-3$
B
)A.$1-\sqrt{3}$
B.$\sqrt{3}-1$
C.$3-\sqrt{3}$
D.$\sqrt{3}-3$
答案：1. B
2. (2024·呼伦贝尔)实数 a,b 在数轴上的对应位置如图所示,则
$\sqrt{(a-b)^{2}}-(b-a-2)$的化简结果是 (
A.2
B.$2a-2$
C.$2-2b$
D.-2
$\sqrt{(a-b)^{2}}-(b-a-2)$的化简结果是 (
A
)A.2
B.$2a-2$
C.$2-2b$
D.-2
答案：2. A
3. 当$a＜\frac{1}{2}$且$a≠0$时,化简$\frac{\sqrt{4a^{2}-4a+1}}{2a^{2}-a}=$
$-\dfrac{1}{a}$
.答案：3. $-\dfrac{1}{a}$
4. 当$1＜x＜5$时,化简:$\sqrt{x^{2}-2x+1}-\sqrt{x^{2}-10x+25}$.
答案：4. 解：$\because 1＜ x＜5$，$\therefore x-1＞0$，$x-5＜0$，
$\therefore$原式$=\sqrt{(x-1)^2}-\sqrt{(x-5)^2}=|x-1|-|x-5|=$
$x-1-(5-x)=2x-6$.
$\therefore$原式$=\sqrt{(x-1)^2}-\sqrt{(x-5)^2}=|x-1|-|x-5|=$
$x-1-(5-x)=2x-6$.
5. 已知$x=5-2\sqrt{6}$,则$x^{2}-10x+1$的值为 (
A.$-30\sqrt{6}$
B.$-18\sqrt{6}-2$
C.0
D.$10\sqrt{6}$
C
)A.$-30\sqrt{6}$
B.$-18\sqrt{6}-2$
C.0
D.$10\sqrt{6}$
答案：5. C
6. 计算:$(\sqrt{3}-\sqrt{2})(-\sqrt{3}-\sqrt{2})+(3+2\sqrt{5})^{2}=$
$28+12\sqrt{5}$
.答案：6. $28+12\sqrt{5}$
7. (2025·秦淮区期末)$(2+\sqrt{5})^{2025}·(2-\sqrt{5})^{2026}=$
$\sqrt{5}-2$
.答案：7. $\sqrt{5}-2$
8. 运用乘法公式计算:
(1)$(2\sqrt{2}+3\sqrt{3})^{2}$; (2)$(\sqrt{3}+2)(2-\sqrt{3})+(\sqrt{3}-\sqrt{2})^{2}$.
(1)$(2\sqrt{2}+3\sqrt{3})^{2}$; (2)$(\sqrt{3}+2)(2-\sqrt{3})+(\sqrt{3}-\sqrt{2})^{2}$.
答案：8. 解：(1)$(2\sqrt{2}+3\sqrt{3})^2=(2\sqrt{2})^2+2×2\sqrt{2}×3\sqrt{3}+(3\sqrt{3})^2$
$=8+12\sqrt{6}+27=35+12\sqrt{6}$.
(2)$(\sqrt{3}+2)(2-\sqrt{3})+(\sqrt{3}-\sqrt{2})^2=4-3+3-2\sqrt{6}+2=$
$6-2\sqrt{6}$.
$=8+12\sqrt{6}+27=35+12\sqrt{6}$.
(2)$(\sqrt{3}+2)(2-\sqrt{3})+(\sqrt{3}-\sqrt{2})^2=4-3+3-2\sqrt{6}+2=$
$6-2\sqrt{6}$.
9. 已知:$a=\sqrt{3}-2,b=\sqrt{3}+2$,求代数式$a^{2}b-ab^{2}$的值.
答案：9. 解：$\because a=\sqrt{3}-2$，$b=\sqrt{3}+2$，
$\therefore a-b=(\sqrt{3}-2)-(\sqrt{3}+2)=-4$，
$ab=(\sqrt{3}-2)×(\sqrt{3}+2)=3-4=-1$，
$\therefore a^2b-ab^2=ab(a-b)=(-1)×(-4)=4$.
$\therefore a-b=(\sqrt{3}-2)-(\sqrt{3}+2)=-4$，
$ab=(\sqrt{3}-2)×(\sqrt{3}+2)=3-4=-1$，
$\therefore a^2b-ab^2=ab(a-b)=(-1)×(-4)=4$.
10. 已知$x=2+\sqrt{3},y=2-\sqrt{3}$,求下列各式的值.
(1)$\frac{1}{x}+\frac{1}{y}$; (2)$\frac{x}{y}+\frac{y}{x}-4$.
(1)$\frac{1}{x}+\frac{1}{y}$; (2)$\frac{x}{y}+\frac{y}{x}-4$.
答案：10. 解：$\because x=2+\sqrt{3}$，$y=2-\sqrt{3}$，$\therefore x+y=4$，$xy=1$.
(1)$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=4$.
(2)$\dfrac{x}{y}+\dfrac{y}{x}-4=\dfrac{x^2+y^2}{xy}-4=\dfrac{x^2+y^2+2xy}{xy}-6$
$=\dfrac{(x+y)^2}{xy}-6=4^2-6=10$.
(1)$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=4$.
(2)$\dfrac{x}{y}+\dfrac{y}{x}-4=\dfrac{x^2+y^2}{xy}-4=\dfrac{x^2+y^2+2xy}{xy}-6$
$=\dfrac{(x+y)^2}{xy}-6=4^2-6=10$.