1. 下列各组数比较大小正确的是(
A.$\sqrt[3]{28}＜3$
B.$\sqrt{24}＞5$
C.$-\sqrt{10}＞-\sqrt{12}$
D.$\frac{10}{3}＜π$
C
).A.$\sqrt[3]{28}＜3$
B.$\sqrt{24}＞5$
C.$-\sqrt{10}＞-\sqrt{12}$
D.$\frac{10}{3}＜π$
答案：A. $\because 3=\sqrt[3]{27},28＞27,\therefore \sqrt[3]{28}＞3,$
$\therefore$此选项错误,故此选项不符合题意;
B. $\because 5=\sqrt{25},24＜25,\therefore \sqrt{24}＜5,$
$\therefore$此选项错误,故此选项不符合题意;
C. $\because |-\sqrt{10}|=\sqrt{10},|-\sqrt{12}|=\sqrt{12},\sqrt{10}＜\sqrt{12},$
$\therefore -\sqrt{10}＞-\sqrt{12},$
$\therefore$此选项正确,故此选项符合题意;
D. $\because \frac{10}{3}\approx 3.33,π\approx 3.14,\therefore \frac{10}{3}＞π,\therefore$此选项错误,故此选项不符合题意. 故选 C.
$\therefore$此选项错误,故此选项不符合题意;
B. $\because 5=\sqrt{25},24＜25,\therefore \sqrt{24}＜5,$
$\therefore$此选项错误,故此选项不符合题意;
C. $\because |-\sqrt{10}|=\sqrt{10},|-\sqrt{12}|=\sqrt{12},\sqrt{10}＜\sqrt{12},$
$\therefore -\sqrt{10}＞-\sqrt{12},$
$\therefore$此选项正确,故此选项符合题意;
D. $\because \frac{10}{3}\approx 3.33,π\approx 3.14,\therefore \frac{10}{3}＞π,\therefore$此选项错误,故此选项不符合题意. 故选 C.
2. 比较$\sqrt{m-5}$和$\sqrt[3]{4-m}$的大小.
答案：根据平方根的定义,可知 $m-5≥0$,即 $m≥5$,
则 $4-m＜0,\sqrt[3]{4-m}＜0.$
又 $\sqrt{m-5}≥0,\therefore \sqrt{m-5}＞\sqrt[3]{4-m}.$
则 $4-m＜0,\sqrt[3]{4-m}＜0.$
又 $\sqrt{m-5}≥0,\therefore \sqrt{m-5}＞\sqrt[3]{4-m}.$
3.已知$a$为有理数，$A=2a^2 - a - 6$，$B=a^2 - a - 10$，则$A$，$B$的大小关系是(
A.$A＞B$
B.$A=B$
C.$A＜B$
D.$A≥ B$
A
)。A.$A＞B$
B.$A=B$
C.$A＜B$
D.$A≥ B$
答案：$\because A=2a^2 -a -6,B=a^2 -a -10,$
$\therefore A-B=(2a^2 -a -6)-(a^2 -a -10)$
$=2a^2 -a -6 -a^2 +a +10=a^2 +4＞0,$
$\therefore A＞B$. 故选 A.
$\therefore A-B=(2a^2 -a -6)-(a^2 -a -10)$
$=2a^2 -a -6 -a^2 +a +10=a^2 +4＞0,$
$\therefore A＞B$. 故选 A.
4. 比较$\frac{\sqrt{3}-1}{6}$与$\frac{1}{6}$的大小.
答案：$\because \frac{\sqrt{3}-1}{6} ÷ \frac{1}{6} =\sqrt{3} -1 ＜ 1,\frac{\sqrt{3}-1}{6} ＞ 0,\frac{1}{6} ＞ 0,$
$\therefore \frac{\sqrt{3}-1}{6}＜\frac{1}{6}.$
$\therefore \frac{\sqrt{3}-1}{6}＜\frac{1}{6}.$
5. (2025·福建漳州期中)阅读材料:像$(\sqrt{13}+3)×(\sqrt{13}-3)=(\sqrt{13})^2 - 3^2 = 4$，$\sqrt{a}·\sqrt{a}=a(a≥0)$，我们称这两个代数式互为有理化因式.在进行二次根式运算时，利用有理化因式可以化去分母中的根号，请你根据上述材料，解决如下问题:
(1)化简:$\frac{3}{\sqrt{6}}=$
(2)①$\sqrt{5}-\sqrt{3}$的有理化因式是
②请利用$\sqrt{5}-\sqrt{3}$的有理化因式化简:
$\frac{2}{\sqrt{5}-\sqrt{3}}=$
(3)比较大小:$\sqrt{2025}-\sqrt{2024}\_\_\_\_\_\_\sqrt{2026}-\sqrt{2025}$.(填“＞”“＜”或“=”)
(1)化简:$\frac{3}{\sqrt{6}}=$
$\frac{\sqrt{6}}{2}$
.(2)①$\sqrt{5}-\sqrt{3}$的有理化因式是
$\sqrt{5}+\sqrt{3}$(答案不唯一)
;②请利用$\sqrt{5}-\sqrt{3}$的有理化因式化简:
$\frac{2}{\sqrt{5}-\sqrt{3}}=$
$\sqrt{5}+\sqrt{3}$
.(3)比较大小:$\sqrt{2025}-\sqrt{2024}\_\_\_\_\_\_\sqrt{2026}-\sqrt{2025}$.(填“＞”“＜”或“=”)
答案：(1)$\frac{\sqrt{6}}{2}$ 解析 $\frac{3}{\sqrt{6}}=\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}.$
(2)①$\sqrt{5}+\sqrt{3}$(答案不唯一) 解析 $\because (\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})=(\sqrt{5})^2-(\sqrt{3})^2=2,$
$\therefore \sqrt{5}-\sqrt{3}$的有理化因式可以是$\sqrt{5}+\sqrt{3}$.(答案不唯一)
②$\sqrt{5}+\sqrt{3}$ 解析 分子、分母同时乘$\sqrt{5}+\sqrt{3}$将分母有理化,即$\frac{2}{\sqrt{5}-\sqrt{3}}=\frac{2(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\sqrt{5}+\sqrt{3}.$
(3)$＞$ 解析 $\because \frac{1}{\sqrt{2025}-\sqrt{2024}}=\sqrt{2025}+\sqrt{2024}＞0,$
$\frac{1}{\sqrt{2026}-\sqrt{2025}}=\sqrt{2026}+\sqrt{2025}＞0,$
且$\sqrt{2026}+\sqrt{2025}＞\sqrt{2025}+\sqrt{2024},$
$\therefore \frac{1}{\sqrt{2026}-\sqrt{2025}}＞\frac{1}{\sqrt{2025}-\sqrt{2024}}.$
又$\sqrt{2026}-\sqrt{2025}＞0,\sqrt{2025}-\sqrt{2024}＞0,$
$\therefore \sqrt{2025}-\sqrt{2024}＞\sqrt{2026}-\sqrt{2025}.$
(2)①$\sqrt{5}+\sqrt{3}$(答案不唯一) 解析 $\because (\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})=(\sqrt{5})^2-(\sqrt{3})^2=2,$
$\therefore \sqrt{5}-\sqrt{3}$的有理化因式可以是$\sqrt{5}+\sqrt{3}$.(答案不唯一)
②$\sqrt{5}+\sqrt{3}$ 解析 分子、分母同时乘$\sqrt{5}+\sqrt{3}$将分母有理化,即$\frac{2}{\sqrt{5}-\sqrt{3}}=\frac{2(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\sqrt{5}+\sqrt{3}.$
(3)$＞$ 解析 $\because \frac{1}{\sqrt{2025}-\sqrt{2024}}=\sqrt{2025}+\sqrt{2024}＞0,$
$\frac{1}{\sqrt{2026}-\sqrt{2025}}=\sqrt{2026}+\sqrt{2025}＞0,$
且$\sqrt{2026}+\sqrt{2025}＞\sqrt{2025}+\sqrt{2024},$
$\therefore \frac{1}{\sqrt{2026}-\sqrt{2025}}＞\frac{1}{\sqrt{2025}-\sqrt{2024}}.$
又$\sqrt{2026}-\sqrt{2025}＞0,\sqrt{2025}-\sqrt{2024}＞0,$
$\therefore \sqrt{2025}-\sqrt{2024}＞\sqrt{2026}-\sqrt{2025}.$
6. (2026·南京玄武区期末)下列整数中，与$\sqrt{22} - 2$最接近的是(
A.1
B.2
C.3
D.4
C
)。A.1
B.2
C.3
D.4
答案：$\because 4^2=16,5^2=25$,而$16＜22＜25$,
$\therefore 4＜\sqrt{22}＜5$. 又$4.5^2=20.25＜22$,
$\therefore 4.5＜\sqrt{22}＜5,\therefore 4.5-2＜\sqrt{22}-2＜5-2$,即$2.5＜\sqrt{22}-2＜3$,$\therefore \sqrt{22}-2$最接近的整数是3. 故选 C.
$\therefore 4＜\sqrt{22}＜5$. 又$4.5^2=20.25＜22$,
$\therefore 4.5＜\sqrt{22}＜5,\therefore 4.5-2＜\sqrt{22}-2＜5-2$,即$2.5＜\sqrt{22}-2＜3$,$\therefore \sqrt{22}-2$最接近的整数是3. 故选 C.
7.实数$a,b$在数轴上的对应点的位置如图所示.把$a,b,-a,-b$按照从小到大的顺序排列,正确的是(
A.$b＜-a＜-b＜a$
B.$-a＜-b＜a＜b$
C.$-b＜a＜-a＜b$
D.$b＜-a＜a＜-b$
D
).A.$b＜-a＜-b＜a$
B.$-a＜-b＜a＜b$
C.$-b＜a＜-a＜b$
D.$b＜-a＜a＜-b$
答案：根据图示,可得$0＜a＜1,b＜-1$,
$\therefore -1＜-a＜0,-b＞1$,
$\therefore b＜-a＜a＜-b$. 故选 D.
$\therefore -1＜-a＜0,-b＞1$,
$\therefore b＜-a＜a＜-b$. 故选 D.
8.（2025·北京丰台区一模）实数$a,b,c$在数轴上的对应点的位置如图所示，下列结论中正确的是（
A.$ac＞0$
B.$b+c＞0$
C.$|c|＞2$
D.$|c|＞|b|$
D
）。A.$ac＞0$
B.$b+c＞0$
C.$|c|＞2$
D.$|c|＞|b|$
答案：D
9.比较$\sqrt{6}+2$和$\sqrt{57}-2$的大小.
答案：$\because 2＜\sqrt{6}＜3,7＜\sqrt{57}＜8$,
$\therefore \sqrt{6}+2＜3+2=5,\sqrt{57}-2＞7-2=5$,
$\therefore \sqrt{6}+2＜\sqrt{57}-2.$
$\therefore \sqrt{6}+2＜3+2=5,\sqrt{57}-2＞7-2=5$,
$\therefore \sqrt{6}+2＜\sqrt{57}-2.$