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7. 计算:
(1)$(\sqrt{5}-\sqrt{2})×(\sqrt{5}+\sqrt{2})$;
(2)$-2\sqrt{3}×3\sqrt{27}$;
(3)$\sqrt{30}×(-\frac{3}{4}\sqrt{2\frac{2}{3}})×\frac{5}{6}\sqrt{\frac{2}{5}}$;
(4)$\sqrt{1\frac{3}{5}}×2\sqrt{3}×(-\frac{1}{2}\sqrt{10})$.
(1)$(\sqrt{5}-\sqrt{2})×(\sqrt{5}+\sqrt{2})$;
(2)$-2\sqrt{3}×3\sqrt{27}$;
(3)$\sqrt{30}×(-\frac{3}{4}\sqrt{2\frac{2}{3}})×\frac{5}{6}\sqrt{\frac{2}{5}}$;
(4)$\sqrt{1\frac{3}{5}}×2\sqrt{3}×(-\frac{1}{2}\sqrt{10})$.
答案
(1) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=(\sqrt{5})^2-(\sqrt{2})^2=5-2=3$
(2) $-2\sqrt{3}×3\sqrt{27}=-6\sqrt{3×27}=-6\sqrt{81}=-6×9=-54$
(3) $\sqrt{30}×(-\frac{3}{4}\sqrt{2\frac{2}{3}})×\frac{5}{6}\sqrt{\frac{2}{5}}$
$=-\frac{3}{4}×\frac{5}{6}×\sqrt{30×\frac{8}{3}×\frac{2}{5}}$
$=-\frac{5}{8}×\sqrt{32}$
$=-\frac{5}{8}×4\sqrt{2}=-\frac{5}{2}\sqrt{2}$
(4) $\sqrt{1\frac{3}{5}}×2\sqrt{3}×(-\frac{1}{2}\sqrt{10})$
$=2×(-\frac{1}{2})×\sqrt{\frac{8}{5}×3×10}$
$=-\sqrt{48}=-4\sqrt{3}$
(2) $-2\sqrt{3}×3\sqrt{27}=-6\sqrt{3×27}=-6\sqrt{81}=-6×9=-54$
(3) $\sqrt{30}×(-\frac{3}{4}\sqrt{2\frac{2}{3}})×\frac{5}{6}\sqrt{\frac{2}{5}}$
$=-\frac{3}{4}×\frac{5}{6}×\sqrt{30×\frac{8}{3}×\frac{2}{5}}$
$=-\frac{5}{8}×\sqrt{32}$
$=-\frac{5}{8}×4\sqrt{2}=-\frac{5}{2}\sqrt{2}$
(4) $\sqrt{1\frac{3}{5}}×2\sqrt{3}×(-\frac{1}{2}\sqrt{10})$
$=2×(-\frac{1}{2})×\sqrt{\frac{8}{5}×3×10}$
$=-\sqrt{48}=-4\sqrt{3}$
解析
(1) $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=(\sqrt{5})^2-(\sqrt{2})^2=5-2=3$
(2) $-2\sqrt{3}×3\sqrt{27}=-6\sqrt{3×27}=-6\sqrt{81}=-6×9=-54$
(3) $\sqrt{30}×(-\frac{3}{4}\sqrt{2\frac{2}{3}})×\frac{5}{6}\sqrt{\frac{2}{5}}$
$=-\frac{3}{4}×\frac{5}{6}×\sqrt{30×\frac{8}{3}×\frac{2}{5}}$
$=-\frac{5}{8}×\sqrt{32}$
$=-\frac{5}{8}×4\sqrt{2}=-\frac{5}{2}\sqrt{2}$
(4) $\sqrt{1\frac{3}{5}}×2\sqrt{3}×(-\frac{1}{2}\sqrt{10})$
$=2×(-\frac{1}{2})×\sqrt{\frac{8}{5}×3×10}$
$=-\sqrt{48}=-4\sqrt{3}$
(2) $-2\sqrt{3}×3\sqrt{27}=-6\sqrt{3×27}=-6\sqrt{81}=-6×9=-54$
(3) $\sqrt{30}×(-\frac{3}{4}\sqrt{2\frac{2}{3}})×\frac{5}{6}\sqrt{\frac{2}{5}}$
$=-\frac{3}{4}×\frac{5}{6}×\sqrt{30×\frac{8}{3}×\frac{2}{5}}$
$=-\frac{5}{8}×\sqrt{32}$
$=-\frac{5}{8}×4\sqrt{2}=-\frac{5}{2}\sqrt{2}$
(4) $\sqrt{1\frac{3}{5}}×2\sqrt{3}×(-\frac{1}{2}\sqrt{10})$
$=2×(-\frac{1}{2})×\sqrt{\frac{8}{5}×3×10}$
$=-\sqrt{48}=-4\sqrt{3}$
8. 定义: 若两个二次根式$a,b$满足$ab = c$, 且$c$是有理数, 则称$a$与$b$是关于$c$的共轭二次根式.
问题解决:
(1) 若$a$与$2\sqrt{3}$是关于$6$的共轭二次根式, 则$a=$
(2) 若$4+\sqrt{3}$与$8-\sqrt{3}m$是关于$26$的共轭二次根式, 求$m$的值.
问题解决:
(1) 若$a$与$2\sqrt{3}$是关于$6$的共轭二次根式, 则$a=$
$\sqrt{3}$
;(2) 若$4+\sqrt{3}$与$8-\sqrt{3}m$是关于$26$的共轭二次根式, 求$m$的值.
答案
8. 解:(1) $\because a$ 与 $2\sqrt{3}$ 是关于 6 的共轭二次根式,$\therefore 2\sqrt{3}a = 6,\therefore a=\frac{6}{2\sqrt{3}}=\sqrt{3},\therefore a=\sqrt{3}$.
(2) $\because 4+\sqrt{3}$ 与 $8-\sqrt{3}m$ 是关于 26 的共轭二次根式,$\therefore (4+\sqrt{3})(8-\sqrt{3}m)=26,\therefore 8-\sqrt{3}m=\frac{26}{4+\sqrt{3}}=\frac{26×(4-\sqrt{3})}{(4+\sqrt{3})×(4-\sqrt{3})}=8 - 2\sqrt{3},\therefore m = 2$.
(2) $\because 4+\sqrt{3}$ 与 $8-\sqrt{3}m$ 是关于 26 的共轭二次根式,$\therefore (4+\sqrt{3})(8-\sqrt{3}m)=26,\therefore 8-\sqrt{3}m=\frac{26}{4+\sqrt{3}}=\frac{26×(4-\sqrt{3})}{(4+\sqrt{3})×(4-\sqrt{3})}=8 - 2\sqrt{3},\therefore m = 2$.
9. (1) 用“$＞$”“$＜$”或“$=$”填空: ①$4 + 3\_\_\_\_\_\_2\sqrt{4×3}$; ②$1+\frac{1}{6}\_\_\_\_\_\_2\sqrt{1×\frac{1}{6}}$; ③$5 + 5\_\_\_\_\_\_2\sqrt{5×5}$;
(2) 由(1)中各式猜想$m + n$与$2\sqrt{mn}(m≥0,n≥0)$的大小, 并说明理由;
(3) 请利用上述结论解决下面问题:
如图, 某园林设计师要对园林的一个区域进行设计改造, 将该区域用篱笆围成矩形的花圃. 花圃恰好可以借用一段墙体, 为了围成面积为$200\ \mathrm{m}^{2}$的花圃, 所用的篱笆至少需要
(2) 由(1)中各式猜想$m + n$与$2\sqrt{mn}(m≥0,n≥0)$的大小, 并说明理由;
(3) 请利用上述结论解决下面问题:
如图, 某园林设计师要对园林的一个区域进行设计改造, 将该区域用篱笆围成矩形的花圃. 花圃恰好可以借用一段墙体, 为了围成面积为$200\ \mathrm{m}^{2}$的花圃, 所用的篱笆至少需要
40
$\mathrm{m}$.答案
9. 解:(1) ① $\because 4 + 3 = 7,2\sqrt{4×3}=4\sqrt{3},\therefore 7^{2}=49$,
$(4\sqrt{3})^{2}=48.\because 49＞48,\therefore 4 + 3＞2\sqrt{4×3}$. ② $\because 1+\frac{1}{6}=\frac{7}{6}＞1,2\sqrt{1×\frac{1}{6}}=\frac{\sqrt{6}}{3}＜1,\therefore 1+\frac{1}{6}＞2\sqrt{1×\frac{1}{6}}$. ③ $\because 5 + 5 = 10,2\sqrt{5×5}=10,\therefore 5 + 5 = 2\sqrt{5×5}$. (2) $m + n≥ 2\sqrt{mn}(m≥ 0,n≥ 0)$.理由:$\because$ 当 $m≥ 0,n≥ 0$ 时,$(\sqrt{m}-\sqrt{n})^{2}≥ 0,\therefore (\sqrt{m})^{2}-2\sqrt{m}·\sqrt{n}+(\sqrt{n})^{2}≥ 0,\therefore m - 2\sqrt{mn}+n≥ 0,\therefore m + n≥ 2\sqrt{mn}$. (3) 设花圃的长为 $a$ m,宽为 $b$ m,则 $a＞0,b＞0,S = ab = 200$,根据(2)的结论,得 $a + 2b≥ 2\sqrt{a·2b}=2\sqrt{2ab}=2\sqrt{2×200}=2×20 = 40$,所以篱笆至少需要 40 m.
$(4\sqrt{3})^{2}=48.\because 49＞48,\therefore 4 + 3＞2\sqrt{4×3}$. ② $\because 1+\frac{1}{6}=\frac{7}{6}＞1,2\sqrt{1×\frac{1}{6}}=\frac{\sqrt{6}}{3}＜1,\therefore 1+\frac{1}{6}＞2\sqrt{1×\frac{1}{6}}$. ③ $\because 5 + 5 = 10,2\sqrt{5×5}=10,\therefore 5 + 5 = 2\sqrt{5×5}$. (2) $m + n≥ 2\sqrt{mn}(m≥ 0,n≥ 0)$.理由:$\because$ 当 $m≥ 0,n≥ 0$ 时,$(\sqrt{m}-\sqrt{n})^{2}≥ 0,\therefore (\sqrt{m})^{2}-2\sqrt{m}·\sqrt{n}+(\sqrt{n})^{2}≥ 0,\therefore m - 2\sqrt{mn}+n≥ 0,\therefore m + n≥ 2\sqrt{mn}$. (3) 设花圃的长为 $a$ m,宽为 $b$ m,则 $a＞0,b＞0,S = ab = 200$,根据(2)的结论,得 $a + 2b≥ 2\sqrt{a·2b}=2\sqrt{2ab}=2\sqrt{2×200}=2×20 = 40$,所以篱笆至少需要 40 m.
