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11 [2024·浙江台州期中]计算:
(1)$\sqrt{8}-\sqrt{3}\div\sqrt{\frac{3}{2}}$;
(2)$(\sqrt{3}+2)(\sqrt{3}-5)+\sqrt{27}$.
(1)$\sqrt{8}-\sqrt{3}\div\sqrt{\frac{3}{2}}$;
(2)$(\sqrt{3}+2)(\sqrt{3}-5)+\sqrt{27}$.
答案：
解：
(1)$\sqrt{8}-\sqrt{3}\div\sqrt{\frac{3}{2}}$
$=2\sqrt{2}-\sqrt{3\times\frac{2}{3}}$
$=2\sqrt{2}-\sqrt{2}$
$=\sqrt{2}$.
(2)$(\sqrt{3}+2)(\sqrt{3}-5)+\sqrt{27}$
$=3-5\sqrt{3}+2\sqrt{3}-10 + 3\sqrt{3}$
$=-7$.
(1)$\sqrt{8}-\sqrt{3}\div\sqrt{\frac{3}{2}}$
$=2\sqrt{2}-\sqrt{3\times\frac{2}{3}}$
$=2\sqrt{2}-\sqrt{2}$
$=\sqrt{2}$.
(2)$(\sqrt{3}+2)(\sqrt{3}-5)+\sqrt{27}$
$=3-5\sqrt{3}+2\sqrt{3}-10 + 3\sqrt{3}$
$=-7$.
12 [2024·浙江杭州期中]下列计算正确的是 ( )
A. $\sqrt{5}\times\sqrt{2}=\sqrt{7}$
B. $\sqrt{(-3)^{2}}=3$
C. $\sqrt{10}\div\sqrt{3}=\sqrt{7}$
D. $5\sqrt{5}-2\sqrt{2}=3\sqrt{3}$
A. $\sqrt{5}\times\sqrt{2}=\sqrt{7}$
B. $\sqrt{(-3)^{2}}=3$
C. $\sqrt{10}\div\sqrt{3}=\sqrt{7}$
D. $5\sqrt{5}-2\sqrt{2}=3\sqrt{3}$
答案：
B
13 新题型 阅读辨析题 下表是嘉嘉和淇淇比较$\sqrt{2}+\sqrt{3}$与$\sqrt{2 + 3}$大小的过程,下列关于两人的思路判断正确的是 ( )
|嘉嘉|淇淇|
|--|--|
|$(\sqrt{2}+\sqrt{3})^{2}=5 + 2\sqrt{6},(\sqrt{2 + 3})^{2}=5.\because5 + 2\sqrt{6}＞5,\therefore\sqrt{2}+\sqrt{3}＞\sqrt{2 + 3}$|作一个两直角边长分别为$\sqrt{2},\sqrt{3}$的直角三角形,利用勾股定理,得斜边长为$\sqrt{(\sqrt{2})^{2}+(\sqrt{3})^{2}}=\sqrt{2 + 3}$.由三角形中两边之和大于第三边,得$\sqrt{2}+\sqrt{3}＞\sqrt{2 + 3}$. |
A. 嘉嘉对,淇淇错
B. 嘉嘉错,淇淇对
C. 两人都对
D. 两人都错
|嘉嘉|淇淇|
|--|--|
|$(\sqrt{2}+\sqrt{3})^{2}=5 + 2\sqrt{6},(\sqrt{2 + 3})^{2}=5.\because5 + 2\sqrt{6}＞5,\therefore\sqrt{2}+\sqrt{3}＞\sqrt{2 + 3}$|作一个两直角边长分别为$\sqrt{2},\sqrt{3}$的直角三角形,利用勾股定理,得斜边长为$\sqrt{(\sqrt{2})^{2}+(\sqrt{3})^{2}}=\sqrt{2 + 3}$.由三角形中两边之和大于第三边,得$\sqrt{2}+\sqrt{3}＞\sqrt{2 + 3}$. |
A. 嘉嘉对,淇淇错
B. 嘉嘉错,淇淇对
C. 两人都对
D. 两人都错
答案：
C
14 新题型 新定义运算题 [2024·浙江杭州期中]对于两个不相等的实数a,b,定义一种新运算:$a※b=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}$,则4※1 =_______.
答案：
3
15 (一题多解)比较二次根式的大小:$-4\sqrt{3}$_______$-3\sqrt{4}$. (填“＞”“=”或“＜”)
答案：
＜ 解析：$\because -4\sqrt{3}=-\sqrt{16}\times\sqrt{3}=-\sqrt{48}$，$-3\sqrt{4}=-\sqrt{9}\times\sqrt{4}=-\sqrt{36}$，$48＞36$，$\therefore -\sqrt{48}＜-\sqrt{36}$，即$-4\sqrt{3}＜-3\sqrt{4}$.
·一题多解（平方比较法）$\because -4\sqrt{3}＜0$，$-3\sqrt{4}＜0$，$(-4\sqrt{3})^{2}=48$，$(-3\sqrt{4})^{2}=36$，$48＞36$，$\therefore (-4\sqrt{3})^{2}＞(-3\sqrt{4})^{2}$，$\therefore -4\sqrt{3}＜-3\sqrt{4}$.
·一题多解（平方比较法）$\because -4\sqrt{3}＜0$，$-3\sqrt{4}＜0$，$(-4\sqrt{3})^{2}=48$，$(-3\sqrt{4})^{2}=36$，$48＞36$，$\therefore (-4\sqrt{3})^{2}＞(-3\sqrt{4})^{2}$，$\therefore -4\sqrt{3}＜-3\sqrt{4}$.
16 [2024·浙江台州期末]计算:
(1)$\sqrt{36}+\sqrt{8}\times\sqrt{\frac{1}{2}}$;
(2)$(\sqrt{2}+1)^{2}-\sqrt{18}\div\sqrt{2}$.
(1)$\sqrt{36}+\sqrt{8}\times\sqrt{\frac{1}{2}}$;
(2)$(\sqrt{2}+1)^{2}-\sqrt{18}\div\sqrt{2}$.
答案：
解：
(1)$\sqrt{36}+\sqrt{8}\times\sqrt{\frac{1}{2}}$
$=6+\sqrt{4}$
$=6 + 2$
$=8$.
(2)$(\sqrt{2}+1)^{2}-\sqrt{18}+\sqrt{2}$
$=2+2\sqrt{2}+1-3\sqrt{2}+\sqrt{2}$
$=3$.
(1)$\sqrt{36}+\sqrt{8}\times\sqrt{\frac{1}{2}}$
$=6+\sqrt{4}$
$=6 + 2$
$=8$.
(2)$(\sqrt{2}+1)^{2}-\sqrt{18}+\sqrt{2}$
$=2+2\sqrt{2}+1-3\sqrt{2}+\sqrt{2}$
$=3$.
17 [2024·浙江杭州月考]若$x=\frac{1}{\sqrt{2}+1},y=\frac{1}{\sqrt{2}-1}$.
(1)求$\frac{1}{x}+\frac{1}{y}$的值;
(2)求$\frac{xy^{2}+x^{2}y}{(x - y)^{2}}$的值.
(1)求$\frac{1}{x}+\frac{1}{y}$的值;
(2)求$\frac{xy^{2}+x^{2}y}{(x - y)^{2}}$的值.
答案：
解：
(1)$\because x=\frac{1}{\sqrt{2}+1}$，$y=\frac{1}{\sqrt{2}-1}$，
$\therefore \frac{1}{x}+\frac{1}{y}=\frac{1}{\frac{1}{\sqrt{2}+1}}+\frac{1}{\frac{1}{\sqrt{2}-1}}=\sqrt{2}+1+\sqrt{2}-1=2\sqrt{2}$.
(2)$\because x=\frac{1}{\sqrt{2}+1}$，$y=\frac{1}{\sqrt{2}-1}$，
$\therefore xy=\frac{1}{2 - 1}=1$，$x + y=\frac{\sqrt{2}-1+\sqrt{2}+1}{2 - 1}=2\sqrt{2}$，$x - y=\frac{\sqrt{2}-1-(\sqrt{2}+1)}{2 - 1}=-2$，
$\therefore \frac{xy^{2}+x^{2}y}{(x - y)^{2}}=\frac{xy(x + y)}{(x - y)^{2}}=\frac{1\times2\sqrt{2}}{(-2)^{2}}=\frac{\sqrt{2}}{2}$.
(1)$\because x=\frac{1}{\sqrt{2}+1}$，$y=\frac{1}{\sqrt{2}-1}$，
$\therefore \frac{1}{x}+\frac{1}{y}=\frac{1}{\frac{1}{\sqrt{2}+1}}+\frac{1}{\frac{1}{\sqrt{2}-1}}=\sqrt{2}+1+\sqrt{2}-1=2\sqrt{2}$.
(2)$\because x=\frac{1}{\sqrt{2}+1}$，$y=\frac{1}{\sqrt{2}-1}$，
$\therefore xy=\frac{1}{2 - 1}=1$，$x + y=\frac{\sqrt{2}-1+\sqrt{2}+1}{2 - 1}=2\sqrt{2}$，$x - y=\frac{\sqrt{2}-1-(\sqrt{2}+1)}{2 - 1}=-2$，
$\therefore \frac{xy^{2}+x^{2}y}{(x - y)^{2}}=\frac{xy(x + y)}{(x - y)^{2}}=\frac{1\times2\sqrt{2}}{(-2)^{2}}=\frac{\sqrt{2}}{2}$.
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