第144页 - 经纶学典学霸八年级数学上下册【江苏国标】

$解:\sqrt{2025}-\sqrt{2023}=(\sqrt{2025}-\sqrt{2023})(\sqrt {2025}+\sqrt {2023})$

$=\frac {2}{\sqrt {2025}+\sqrt {2023}}$

$同理 \sqrt{2024}- \sqrt{2022}=\frac{2}{\sqrt{2024}+\sqrt {2022}}$

$∵ \sqrt{2025}+ \sqrt{2023}＞ \sqrt{2024}+ \sqrt{2022}$

$∴ \sqrt {2025}- \sqrt{2023}＜\sqrt{2024}-\sqrt{2022}$

$解:3\sqrt {2}-4= \frac{2}{3\sqrt {2}+4}$

$2\sqrt {3}-\sqrt {10}=\frac{2}{2\sqrt {3}+\sqrt {10}}$

$而3\sqrt {2}＞2\sqrt {3},4＞ \sqrt{10}$

$∴3\sqrt {2}+4＞2\sqrt {3}+\sqrt {10}$

$∴3\sqrt {2}-4＜2\sqrt {3}- \sqrt{10}$

$解:通过移项可得比较原式的大小等同于比较$

$\sqrt{a+7}- \sqrt{a+5}与 \sqrt{a+5}- \sqrt{a+3}的大小$

$\sqrt{a+7}-\sqrt{a+5}=\frac{2}{\sqrt{a+7}+\sqrt{a+5}}$

$\ \sqrt{a+5}-\sqrt{a+3}=\frac{2}{\sqrt{a+5}+\sqrt{a+3}}$

$∵ \sqrt{a+7}+ \sqrt{a+5}＞\sqrt{a+5}+ \sqrt{a+3}$

$∴ \sqrt{a+7}+\sqrt {a+5}＜ \sqrt{a+5}-\sqrt {a+3}$

$∴ \sqrt{a+3}+\sqrt{a+7}＜2 \sqrt{a+5}，即P＜Q$

$解:∵2x+1≥0,2x-3≥0,∴x≥\frac{3}{2}$

$∵y= \sqrt{2x+1}-\sqrt {2x-3}+5=\frac{4}{\sqrt{2x+1}+\sqrt{2x-3}}+5$

$当x=\frac{3}{2}时，分母\sqrt {2x+1}+\sqrt {2x-3}有最小值2$

$∴y=\frac{4}{\sqrt {2x+1}+\sqrt{2x-3}}+5有最大值7$

$解:由1-x≥0,1+x≥0,x≥0得0≤x≤1$

$y= \sqrt{1-x}+\sqrt{1+x}-\sqrt{x}=$

$\ \sqrt{1-x}+\frac {1}{\sqrt{1+x}+\sqrt{x}}$

$当x=1时， \sqrt{1+x}+\sqrt{x}有最大值$

$则\frac{1}{\sqrt{1+x}+\sqrt{x}}有最小值\sqrt {2}-1$

$此时\sqrt{1-x}有最小值0$

$∴y的最小值\sqrt {2}-1$