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14. (2024·淮阴区期末)计算:(1)$-1^{2024}-|1-\sqrt{2}|+(π-2)^0$;
(2)$-\sqrt[3]{27}+(-2)^2+\sqrt{16}÷(-\dfrac{2}{3})$;
(3)$|-\sqrt{2}|-(\sqrt{3}-\sqrt{2})-|\sqrt{3}-2|$;
(4)$\sqrt[3]{125}+\sqrt{(-2)^2}-\sqrt{|-25|}-(-1)^{2024}$.
(2)$-\sqrt[3]{27}+(-2)^2+\sqrt{16}÷(-\dfrac{2}{3})$;
(3)$|-\sqrt{2}|-(\sqrt{3}-\sqrt{2})-|\sqrt{3}-2|$;
(4)$\sqrt[3]{125}+\sqrt{(-2)^2}-\sqrt{|-25|}-(-1)^{2024}$.
答案
14.解:(1)原式$=-1-(\sqrt{2}-1)+1=1-\sqrt{2}$.
(2)原式$=-3+4+4×(-\dfrac{3}{2})=1-6=-5$.
(3)原式$=\sqrt{2}-\sqrt{3}+\sqrt{2}-(2-\sqrt{3})=\sqrt{2}-\sqrt{3}+\sqrt{2}-2+\sqrt{3}=2\sqrt{2}-2$.
(4)原式$=5+2-5-1=1$.
(2)原式$=-3+4+4×(-\dfrac{3}{2})=1-6=-5$.
(3)原式$=\sqrt{2}-\sqrt{3}+\sqrt{2}-(2-\sqrt{3})=\sqrt{2}-\sqrt{3}+\sqrt{2}-2+\sqrt{3}=2\sqrt{2}-2$.
(4)原式$=5+2-5-1=1$.
15. 阅读材料,并回答问题:
$\because 4＜5＜9,\therefore \sqrt{4}＜\sqrt{5}＜\sqrt{9}$,即$2＜\sqrt{5}＜3,\therefore \sqrt{5}$的整数部分为 2,小数部分为$\sqrt{5}-2$.
(1)类比上述方法,求$\sqrt{20}$的整数部分和小数部分;
(2)试判断$\dfrac{\sqrt{20}-4}{3}$与$\dfrac{1}{3}$的大小关系,并说明理由.
$\because 4＜5＜9,\therefore \sqrt{4}＜\sqrt{5}＜\sqrt{9}$,即$2＜\sqrt{5}＜3,\therefore \sqrt{5}$的整数部分为 2,小数部分为$\sqrt{5}-2$.
(1)类比上述方法,求$\sqrt{20}$的整数部分和小数部分;
(2)试判断$\dfrac{\sqrt{20}-4}{3}$与$\dfrac{1}{3}$的大小关系,并说明理由.
答案
15.解:(1)$\because 16＜20＜25,\therefore 4＜\sqrt{20}＜5$,
则$\sqrt{20}$的整数部分为 4,小数部分为$\sqrt{20}-4$.
(2)$\dfrac{\sqrt{20}-4}{3}＜\dfrac{1}{3}$,理由如下：
$\because \dfrac{\sqrt{20}-4}{3}-\dfrac{1}{3}=\dfrac{\sqrt{20}-5}{3}＜0,\therefore \dfrac{\sqrt{20}-4}{3}＜\dfrac{1}{3}$.
则$\sqrt{20}$的整数部分为 4,小数部分为$\sqrt{20}-4$.
(2)$\dfrac{\sqrt{20}-4}{3}＜\dfrac{1}{3}$,理由如下：
$\because \dfrac{\sqrt{20}-4}{3}-\dfrac{1}{3}=\dfrac{\sqrt{20}-5}{3}＜0,\therefore \dfrac{\sqrt{20}-4}{3}＜\dfrac{1}{3}$.
16. 如图,实数 $a,b,c$ 是数轴上三点 $A,B,C$ 所对应的数,化简: $\sqrt{a^2}+|a-b|+\sqrt[3]{(a+b)^3}-|b-c|$.
答案
16.解:原式$=|a|+|a-b|+a+b-|b-c|=-a+a-b+a+b-c+b=a+b-c$.
17. 比较两个实数的大小,有多种方法.
例如,比较$\dfrac{\sqrt{3}-1}{3}$与$\dfrac{1}{3}$的大小.
方法一:$\dfrac{\sqrt{3}-1}{3}-\dfrac{1}{3}=\dfrac{\sqrt{3}-2}{3}$.
$\because \sqrt{3}-2＜0,\therefore \dfrac{\sqrt{3}-1}{3}-\dfrac{1}{3}＜0$,即$\dfrac{\sqrt{3}-1}{3}＜\dfrac{1}{3}$.
方法二:$\because \dfrac{\sqrt{3}-1}{3}\approx 0.244,0.244＜\dfrac{1}{3},\therefore \dfrac{\sqrt{3}-1}{3}＜\dfrac{1}{3}$.
用两种方法比较$\sqrt{7}+5$与$11-\sqrt{7}$的大小.$(\sqrt{7}\approx 2.646)$
例如,比较$\dfrac{\sqrt{3}-1}{3}$与$\dfrac{1}{3}$的大小.
方法一:$\dfrac{\sqrt{3}-1}{3}-\dfrac{1}{3}=\dfrac{\sqrt{3}-2}{3}$.
$\because \sqrt{3}-2＜0,\therefore \dfrac{\sqrt{3}-1}{3}-\dfrac{1}{3}＜0$,即$\dfrac{\sqrt{3}-1}{3}＜\dfrac{1}{3}$.
方法二:$\because \dfrac{\sqrt{3}-1}{3}\approx 0.244,0.244＜\dfrac{1}{3},\therefore \dfrac{\sqrt{3}-1}{3}＜\dfrac{1}{3}$.
用两种方法比较$\sqrt{7}+5$与$11-\sqrt{7}$的大小.$(\sqrt{7}\approx 2.646)$
答案
17.解:方法一:$\sqrt{7}+5-(11-\sqrt{7})=\sqrt{7}+5-11+\sqrt{7}=2\sqrt{7}-6$.
$\because \sqrt{7}＜3,\therefore 2\sqrt{7}＜6,\therefore 2\sqrt{7}-6＜0,\therefore \sqrt{7}+5＜11-\sqrt{7}$.
方法二:$\because \sqrt{7}\approx 2.646,\sqrt{7}+5\approx 2.646+5=7.646$,
$11-\sqrt{7}\approx 11-2.646=8.354,7.646＜8.354$,
$\therefore \sqrt{7}+5＜11-\sqrt{7}$.
$\because \sqrt{7}＜3,\therefore 2\sqrt{7}＜6,\therefore 2\sqrt{7}-6＜0,\therefore \sqrt{7}+5＜11-\sqrt{7}$.
方法二:$\because \sqrt{7}\approx 2.646,\sqrt{7}+5\approx 2.646+5=7.646$,
$11-\sqrt{7}\approx 11-2.646=8.354,7.646＜8.354$,
$\therefore \sqrt{7}+5＜11-\sqrt{7}$.
