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1.「2026江苏常州期中」比较大小：
$\sqrt[3]{8}$
$\sqrt[3]{8}$
＜
$\sqrt{5}$，$\sqrt{7}$ ＞
$\frac{5}{2}$答案
答案 ＜;＞
解析 $\because \sqrt[3]{8}=2,2^{2}=4,(\sqrt{5})^{2}=5,4＜5,\therefore \sqrt[3]{8}＜\sqrt{5}.$
$\because (\sqrt{7})^{2}=7,(\dfrac{5}{2})^{2}=\dfrac{25}{4},7=\dfrac{28}{4}＞\dfrac{25}{4},\therefore \sqrt{7}＞\dfrac{5}{2}.$
解析 $\because \sqrt[3]{8}=2,2^{2}=4,(\sqrt{5})^{2}=5,4＜5,\therefore \sqrt[3]{8}＜\sqrt{5}.$
$\because (\sqrt{7})^{2}=7,(\dfrac{5}{2})^{2}=\dfrac{25}{4},7=\dfrac{28}{4}＞\dfrac{25}{4},\therefore \sqrt{7}＞\dfrac{5}{2}.$
2.「2026江苏泰州姜堰期中」已知$\sqrt[3]{a}=1.2$，$\sqrt{b}=1.2$，那么a,b的大小关系是
$a＞b$
。答案
答案 $a＞b$
解析 $\because \sqrt[3]{a}=1.2,\sqrt{b}=1.2,\therefore a=1.2^{3}=1.728,b=1.2^{2}=1.44,\because 1.728＞1.44,\therefore a＞b.$
解析 $\because \sqrt[3]{a}=1.2,\sqrt{b}=1.2,\therefore a=1.2^{3}=1.728,b=1.2^{2}=1.44,\because 1.728＞1.44,\therefore a＞b.$
3.比较大小：$-\dfrac{\sqrt{5}}{2}$与$-\dfrac{\sqrt{10}}{3}$。
答案
解析 $(-\dfrac{\sqrt{5}}{2})^{2}=\dfrac{5}{4}=\dfrac{45}{36},(-\dfrac{\sqrt{10}}{3})^{2}=\dfrac{10}{9}=\dfrac{40}{36},$
$\because \dfrac{45}{36}＞\dfrac{40}{36},\therefore \dfrac{\sqrt{5}}{2}＞\dfrac{\sqrt{10}}{3},\therefore -\dfrac{\sqrt{5}}{2}＜-\dfrac{\sqrt{10}}{3}.$
$\because \dfrac{45}{36}＞\dfrac{40}{36},\therefore \dfrac{\sqrt{5}}{2}＞\dfrac{\sqrt{10}}{3},\therefore -\dfrac{\sqrt{5}}{2}＜-\dfrac{\sqrt{10}}{3}.$
4.(1)比较16与$\sqrt{255}$的大小.
(2)比较$\sqrt[3]{-25}$与-3的大小.
(2)比较$\sqrt[3]{-25}$与-3的大小.
答案
解析 (1)$\because 255＜256,\therefore \sqrt{255}＜\sqrt{256},即\sqrt{255}＜16.$
(2)$\because -25＞-27,\therefore \sqrt[3]{-25}＞\sqrt[3]{-27},\therefore \sqrt[3]{-25}＞-3.$
(2)$\because -25＞-27,\therefore \sqrt[3]{-25}＞\sqrt[3]{-27},\therefore \sqrt[3]{-25}＞-3.$
5.「2026 江苏宿迁宿城期中」比较大小：$\sqrt{5}+1$
＞
3.答案
答案 ＞
解析 $\because 2＜\sqrt{5}＜3,\therefore \sqrt{5}+1＞3.$
解析 $\because 2＜\sqrt{5}＜3,\therefore \sqrt{5}+1＞3.$
6. 比较大小：$-\sqrt{3}$
＞
$-π.$答案
答案 ＞
解析 $\because -\sqrt{3}\approx -1.732,-π\approx -3.14,\therefore -\sqrt{3}＞-π.$
解析 $\because -\sqrt{3}\approx -1.732,-π\approx -3.14,\therefore -\sqrt{3}＞-π.$
7.比较$\frac{\sqrt{5}-2}{2}$与0.5的大小.
答案
解析 $\because \dfrac{\sqrt{5}-2}{2}-0.5=\dfrac{\sqrt{5}-2}{2}-\dfrac{1}{2}=\dfrac{\sqrt{5}-3}{2}＜0,$
$\therefore \dfrac{\sqrt{5}-2}{2}＜0.5.$
$\therefore \dfrac{\sqrt{5}-2}{2}＜0.5.$
8.比较$\frac{\sqrt{2}+1}{6}$与$\frac{2}{3}$的大小.
答案
解析 $\dfrac{\sqrt{2}+1}{6}-\dfrac{2}{3}=\dfrac{\sqrt{2}+1-4}{6}=\dfrac{\sqrt{2}-3}{6},$
$\because \sqrt{2}＜3,\therefore \sqrt{2}-3＜0,\therefore \dfrac{\sqrt{2}+1}{6}-\dfrac{2}{3}＜0,\therefore \dfrac{\sqrt{2}+1}{6}＜\dfrac{2}{3}.$
$\because \sqrt{2}＜3,\therefore \sqrt{2}-3＜0,\therefore \dfrac{\sqrt{2}+1}{6}-\dfrac{2}{3}＜0,\therefore \dfrac{\sqrt{2}+1}{6}＜\dfrac{2}{3}.$
9.比较大小：$\frac{2026}{2025}$与$\frac{2025}{2024}$。
答案
解析 $\dfrac{2\ 026}{2\ 025}÷ \dfrac{2\ 025}{2\ 024}=\dfrac{2\ 026}{2\ 025}× \dfrac{2\ 024}{2\ 025}=$
$\dfrac{(2\ 025+1)×(2\ 025-1)}{2\ 025^{2}}=\dfrac{2\ 025^{2}-1}{2\ 025^{2}}=1-\dfrac{1}{2\ 025^{2}}＜1,$
$\therefore \dfrac{2\ 026}{2\ 025}＜\dfrac{2\ 025}{2\ 024}.$
$\dfrac{(2\ 025+1)×(2\ 025-1)}{2\ 025^{2}}=\dfrac{2\ 025^{2}-1}{2\ 025^{2}}=1-\dfrac{1}{2\ 025^{2}}＜1,$
$\therefore \dfrac{2\ 026}{2\ 025}＜\dfrac{2\ 025}{2\ 024}.$
