第148页

信息发布者:
$\sqrt{5}\pm 2$
解:​$(1)$​
$\begin{aligned} 原式&=3\sqrt{2}-\frac{3\sqrt{2}}{2}+2\sqrt{2}\\ &=\frac{6\sqrt{2}}{2}-\frac{3\sqrt{2}}{2}+\frac{4\sqrt{2}}{2}\\ &=\frac{7\sqrt{2}}{2} \end{aligned}$
解:​$(2)$​
$\begin{aligned} 原式&=\sqrt{3a^2}×\sqrt{\frac{2}{9a}}×\frac{1}{2}\sqrt{\frac{2a}{3}}\\ &=\sqrt{3a^2·\frac{2}{9a}·\frac{2a}{3}}\\ &=\sqrt{\frac{4a^2}{9}}\\ &=\frac{a}{3} \end{aligned}$
解:​$(3)$​
$\begin{aligned} 原式&=(\sqrt{5})^2-(\sqrt{3})^2-((2\sqrt{3})^2 - 2×2\sqrt{3}×\sqrt{2}+(\sqrt{2})^2)\\ &=5-3-(12-4\sqrt{6}+2)\\ &=2-14+4\sqrt{6}\\ &=4\sqrt{6}-12 \end{aligned}$
解:​$(4)$​
$\begin{aligned} 原式&=-1+4-\sqrt{\frac{1}{3}×48}+\sqrt{2}-1+1\\ &=-1+4-4+\sqrt{2}-1+1\\ &=\sqrt{2}-1 \end{aligned}$
解:$(1)$设长方体的长为$4x$,宽为$2x$,高为$\sqrt {5}x$,
由题意得:
$ 4x·2x=20$,
$ 8x^2=20$,
$ x^2=\frac {5}{2}$,
∵$x>0$,
∴$x=\frac {\sqrt {10}}{2}$,
$ $则高为$\sqrt {5}×\frac {\sqrt {10}}{2}=\frac {5\sqrt {2}}{2}$。
$ (2) $长为$4×\frac {\sqrt {10}}{2}=2\sqrt {10}$,宽为$2×\frac {\sqrt {10}}{2}=\sqrt {10}$,
表面积=2×(长$×$宽$+$长$×$高$+$宽$×$高)
​$=2×(20+2\sqrt {10}×\frac {5\sqrt {2}}{2}+\sqrt {10}×\frac {5\sqrt {2}}{2})$​
​$=2×(20+5\sqrt {20}+\frac {5\sqrt {20}}{2})$​
​$=2×(20+10\sqrt {5}+5\sqrt {5})$​
​$=2×(20+15\sqrt {5})$​
​$=40+30\sqrt {5} $​
解:​$(1)$​
$\begin{aligned} 原式&=\frac{2}{3}×3\sqrt{x}+6×\frac{\sqrt{x}}{2}-2x·\frac{\sqrt{x}}{x}\\ &=2\sqrt{x}+3\sqrt{x}-2\sqrt{x}\\ &=3\sqrt{x} \end{aligned}$
取$x=4$($x>0$即可),则原式$=3\sqrt{4}=6。$
解:$(2)$∵$a=\frac {1}{\sqrt {3}+\sqrt {2}}=\sqrt {3}-\sqrt {2}$,
∴$a-1=\sqrt {3}-\sqrt {2}-1<0$,
​$ $​原式​$=\frac {(a-1)^2}{a-1}-\frac {\sqrt {(a-1)^2}}{a(a-1)}$​
​$=a-1-\frac {1-a}{a(a-1)}$​
​$=a-1+\frac {1}{a}$​
∵$\frac {1}{a}=\sqrt {3}+\sqrt {2}$,
∴原式$=\sqrt {3}-\sqrt {2}-1+\sqrt {3}+\sqrt {2}=2\sqrt {3}-1$
解:(1)$∵$ 点$A$表示$\sqrt{3},$点$B$表示$\sqrt{5},$
$∴AB=\sqrt{5}-\sqrt{3},$
$∵$ 点$C$是点$B$关于点$A$的对称点,
$∴AC=AB,$即$\sqrt{3}-x=\sqrt{5}-\sqrt{3},$
解得$x=2\sqrt{3}-\sqrt{5}。$
(2) $∵x=2\sqrt{3}-\sqrt{5},$
$∴x^2=(2\sqrt{3}-\sqrt{5})^2=12-4\sqrt{15}+5=17-4\sqrt{15},$
$\begin{aligned} 原式&=(17+4\sqrt{15})(17-4\sqrt{15})-(2\sqrt{3}+\sqrt{5})(2\sqrt{3}-\sqrt{5})-2\\ &=17^2-(4\sqrt{15})^2-[(2\sqrt{3})^2-(\sqrt{5})^2]-2\\ &=289-240-(12-5)-2\\ &=49-7-2\\ &=40 \end{aligned}$