登录
5. 计算：
(1) $(-\frac{1}{2}+\frac{\sqrt{3}}{2})^{2}$；
(2) $(\sqrt{24}-3\sqrt{15}+2\sqrt{2\frac{2}{3}})×\sqrt{2}$；
(3) $4\sqrt{\frac{9}{8}}×\frac{1}{2}\sqrt{\frac{49}{50}}-\sqrt{\frac{9}{28}}÷\sqrt{1\frac{1}{35}}$；
(4) $\sqrt{45}÷2\sqrt{\frac{1}{5}}×\sqrt{2\frac{2}{3}}-2\sqrt{3}×(\frac{3}{2}\sqrt{2}-3\sqrt{\frac{3}{8}})$；
(5) $(x+\sqrt{y})(x-\sqrt{y})-(2x-\sqrt{y})(x + 2\sqrt{y})$；
(6) $(1 - 2\sqrt{3})(1 + 2\sqrt{3})-(2\sqrt{3}-1)^{2}$。
(1) $(-\frac{1}{2}+\frac{\sqrt{3}}{2})^{2}$；
(2) $(\sqrt{24}-3\sqrt{15}+2\sqrt{2\frac{2}{3}})×\sqrt{2}$；
(3) $4\sqrt{\frac{9}{8}}×\frac{1}{2}\sqrt{\frac{49}{50}}-\sqrt{\frac{9}{28}}÷\sqrt{1\frac{1}{35}}$；
(4) $\sqrt{45}÷2\sqrt{\frac{1}{5}}×\sqrt{2\frac{2}{3}}-2\sqrt{3}×(\frac{3}{2}\sqrt{2}-3\sqrt{\frac{3}{8}})$；
(5) $(x+\sqrt{y})(x-\sqrt{y})-(2x-\sqrt{y})(x + 2\sqrt{y})$；
(6) $(1 - 2\sqrt{3})(1 + 2\sqrt{3})-(2\sqrt{3}-1)^{2}$。
答案
(1) $(-\frac{1}{2}+\frac{\sqrt{3}}{2})^{2}$
$=(-\frac{1}{2})^{2}+2×(-\frac{1}{2})×\frac{\sqrt{3}}{2}+(\frac{\sqrt{3}}{2})^{2}$
$=\frac{1}{4}-\frac{\sqrt{3}}{2}+\frac{3}{4}$
$=1 - \frac{\sqrt{3}}{2}$
(2) $(\sqrt{24}-3\sqrt{15}+2\sqrt{2\frac{2}{3}})×\sqrt{2}$
$=(2\sqrt{6}-3\sqrt{15}+2×\frac{2\sqrt{6}}{3})×\sqrt{2}$
$=(2\sqrt{6}-3\sqrt{15}+\frac{4\sqrt{6}}{3})×\sqrt{2}$
$=(\frac{10\sqrt{6}}{3}-3\sqrt{15})×\sqrt{2}$
$=\frac{10\sqrt{12}}{3}-3\sqrt{30}$
$=\frac{20\sqrt{3}}{3}-3\sqrt{30}$
$=\frac{20}{3}\sqrt{3} - 3\sqrt{30}$
(3) $4\sqrt{\frac{9}{8}}×\frac{1}{2}\sqrt{\frac{49}{50}}-\sqrt{\frac{9}{28}}÷\sqrt{1\frac{1}{35}}$
$=4×\frac{3}{2\sqrt{2}}×\frac{1}{2}×\frac{7}{5\sqrt{2}}-\sqrt{\frac{9}{28}÷\frac{36}{35}}$
$=4×\frac{3}{2\sqrt{2}}×\frac{7}{10\sqrt{2}}-\sqrt{\frac{9}{28}×\frac{35}{36}}$
$=\frac{84}{40}-\sqrt{\frac{5}{16}}$
$=\frac{21}{10}-\frac{\sqrt{5}}{4}$
(4) $\sqrt{45}÷2\sqrt{\frac{1}{5}}×\sqrt{2\frac{2}{3}}-2\sqrt{3}×(\frac{3}{2}\sqrt{2}-3\sqrt{\frac{3}{8}})$
$=3\sqrt{5}÷\frac{2}{\sqrt{5}}×\frac{2\sqrt{6}}{3}-2\sqrt{3}×(\frac{3}{2}\sqrt{2}-\frac{3\sqrt{6}}{4})$
$=3\sqrt{5}×\frac{\sqrt{5}}{2}×\frac{2\sqrt{6}}{3}-3\sqrt{6}+\frac{3\sqrt{18}}{2}$
$=5\sqrt{6}-3\sqrt{6}+\frac{9\sqrt{2}}{2}$
$=2\sqrt{6}+\frac{9}{2}\sqrt{2}$
(5) $(x+\sqrt{y})(x-\sqrt{y})-(2x-\sqrt{y})(x + 2\sqrt{y})$
$=x^{2}-y-(2x^{2}+4x\sqrt{y}-x\sqrt{y}-2y)$
$=x^{2}-y-2x^{2}-3x\sqrt{y}+2y$
$=-x^{2}+y - 3x\sqrt{y}$
(6) $(1 - 2\sqrt{3})(1 + 2\sqrt{3})-(2\sqrt{3}-1)^{2}$
$=1-(2\sqrt{3})^{2}-[(2\sqrt{3})^{2}-4\sqrt{3}+1]$
$=1 - 12 - (12 - 4\sqrt{3}+1)$
$=-11 - 13 + 4\sqrt{3}$
$=-24 + 4\sqrt{3}$
$=(-\frac{1}{2})^{2}+2×(-\frac{1}{2})×\frac{\sqrt{3}}{2}+(\frac{\sqrt{3}}{2})^{2}$
$=\frac{1}{4}-\frac{\sqrt{3}}{2}+\frac{3}{4}$
$=1 - \frac{\sqrt{3}}{2}$
(2) $(\sqrt{24}-3\sqrt{15}+2\sqrt{2\frac{2}{3}})×\sqrt{2}$
$=(2\sqrt{6}-3\sqrt{15}+2×\frac{2\sqrt{6}}{3})×\sqrt{2}$
$=(2\sqrt{6}-3\sqrt{15}+\frac{4\sqrt{6}}{3})×\sqrt{2}$
$=(\frac{10\sqrt{6}}{3}-3\sqrt{15})×\sqrt{2}$
$=\frac{10\sqrt{12}}{3}-3\sqrt{30}$
$=\frac{20\sqrt{3}}{3}-3\sqrt{30}$
$=\frac{20}{3}\sqrt{3} - 3\sqrt{30}$
(3) $4\sqrt{\frac{9}{8}}×\frac{1}{2}\sqrt{\frac{49}{50}}-\sqrt{\frac{9}{28}}÷\sqrt{1\frac{1}{35}}$
$=4×\frac{3}{2\sqrt{2}}×\frac{1}{2}×\frac{7}{5\sqrt{2}}-\sqrt{\frac{9}{28}÷\frac{36}{35}}$
$=4×\frac{3}{2\sqrt{2}}×\frac{7}{10\sqrt{2}}-\sqrt{\frac{9}{28}×\frac{35}{36}}$
$=\frac{84}{40}-\sqrt{\frac{5}{16}}$
$=\frac{21}{10}-\frac{\sqrt{5}}{4}$
(4) $\sqrt{45}÷2\sqrt{\frac{1}{5}}×\sqrt{2\frac{2}{3}}-2\sqrt{3}×(\frac{3}{2}\sqrt{2}-3\sqrt{\frac{3}{8}})$
$=3\sqrt{5}÷\frac{2}{\sqrt{5}}×\frac{2\sqrt{6}}{3}-2\sqrt{3}×(\frac{3}{2}\sqrt{2}-\frac{3\sqrt{6}}{4})$
$=3\sqrt{5}×\frac{\sqrt{5}}{2}×\frac{2\sqrt{6}}{3}-3\sqrt{6}+\frac{3\sqrt{18}}{2}$
$=5\sqrt{6}-3\sqrt{6}+\frac{9\sqrt{2}}{2}$
$=2\sqrt{6}+\frac{9}{2}\sqrt{2}$
(5) $(x+\sqrt{y})(x-\sqrt{y})-(2x-\sqrt{y})(x + 2\sqrt{y})$
$=x^{2}-y-(2x^{2}+4x\sqrt{y}-x\sqrt{y}-2y)$
$=x^{2}-y-2x^{2}-3x\sqrt{y}+2y$
$=-x^{2}+y - 3x\sqrt{y}$
(6) $(1 - 2\sqrt{3})(1 + 2\sqrt{3})-(2\sqrt{3}-1)^{2}$
$=1-(2\sqrt{3})^{2}-[(2\sqrt{3})^{2}-4\sqrt{3}+1]$
$=1 - 12 - (12 - 4\sqrt{3}+1)$
$=-11 - 13 + 4\sqrt{3}$
$=-24 + 4\sqrt{3}$
解析
(1) $(-\frac{1}{2}+\frac{\sqrt{3}}{2})^{2}$
$=(-\frac{1}{2})^{2}+2×(-\frac{1}{2})×\frac{\sqrt{3}}{2}+(\frac{\sqrt{3}}{2})^{2}$
$=\frac{1}{4}-\frac{\sqrt{3}}{2}+\frac{3}{4}$
$=1 - \frac{\sqrt{3}}{2}$
(2) $(\sqrt{24}-3\sqrt{15}+2\sqrt{2\frac{2}{3}})×\sqrt{2}$
$=(2\sqrt{6}-3\sqrt{15}+2×\frac{2\sqrt{6}}{3})×\sqrt{2}$
$=(2\sqrt{6}-3\sqrt{15}+\frac{4\sqrt{6}}{3})×\sqrt{2}$
$=(\frac{10\sqrt{6}}{3}-3\sqrt{15})×\sqrt{2}$
$=\frac{10\sqrt{12}}{3}-3\sqrt{30}$
$=\frac{20\sqrt{3}}{3}-3\sqrt{30}$
$=\frac{20}{3}\sqrt{3} - 3\sqrt{30}$
(3) $4\sqrt{\frac{9}{8}}×\frac{1}{2}\sqrt{\frac{49}{50}}-\sqrt{\frac{9}{28}}÷\sqrt{1\frac{1}{35}}$
$=4×\frac{3}{2\sqrt{2}}×\frac{1}{2}×\frac{7}{5\sqrt{2}}-\sqrt{\frac{9}{28}÷\frac{36}{35}}$
$=4×\frac{3}{2\sqrt{2}}×\frac{7}{10\sqrt{2}}-\sqrt{\frac{9}{28}×\frac{35}{36}}$
$=\frac{84}{40}-\sqrt{\frac{5}{16}}$
$=\frac{21}{10}-\frac{\sqrt{5}}{4}$
(4) $\sqrt{45}÷2\sqrt{\frac{1}{5}}×\sqrt{2\frac{2}{3}}-2\sqrt{3}×(\frac{3}{2}\sqrt{2}-3\sqrt{\frac{3}{8}})$
$=3\sqrt{5}÷\frac{2}{\sqrt{5}}×\frac{2\sqrt{6}}{3}-2\sqrt{3}×(\frac{3}{2}\sqrt{2}-\frac{3\sqrt{6}}{4})$
$=3\sqrt{5}×\frac{\sqrt{5}}{2}×\frac{2\sqrt{6}}{3}-3\sqrt{6}+\frac{3\sqrt{18}}{2}$
$=5\sqrt{6}-3\sqrt{6}+\frac{9\sqrt{2}}{2}$
$=2\sqrt{6}+\frac{9}{2}\sqrt{2}$
(5) $(x+\sqrt{y})(x-\sqrt{y})-(2x-\sqrt{y})(x + 2\sqrt{y})$
$=x^{2}-y-(2x^{2}+4x\sqrt{y}-x\sqrt{y}-2y)$
$=x^{2}-y-2x^{2}-3x\sqrt{y}+2y$
$=-x^{2}+y - 3x\sqrt{y}$
(6) $(1 - 2\sqrt{3})(1 + 2\sqrt{3})-(2\sqrt{3}-1)^{2}$
$=1-(2\sqrt{3})^{2}-[(2\sqrt{3})^{2}-4\sqrt{3}+1]$
$=1 - 12 - (12 - 4\sqrt{3}+1)$
$=-11 - 13 + 4\sqrt{3}$
$=-24 + 4\sqrt{3}$
$=(-\frac{1}{2})^{2}+2×(-\frac{1}{2})×\frac{\sqrt{3}}{2}+(\frac{\sqrt{3}}{2})^{2}$
$=\frac{1}{4}-\frac{\sqrt{3}}{2}+\frac{3}{4}$
$=1 - \frac{\sqrt{3}}{2}$
(2) $(\sqrt{24}-3\sqrt{15}+2\sqrt{2\frac{2}{3}})×\sqrt{2}$
$=(2\sqrt{6}-3\sqrt{15}+2×\frac{2\sqrt{6}}{3})×\sqrt{2}$
$=(2\sqrt{6}-3\sqrt{15}+\frac{4\sqrt{6}}{3})×\sqrt{2}$
$=(\frac{10\sqrt{6}}{3}-3\sqrt{15})×\sqrt{2}$
$=\frac{10\sqrt{12}}{3}-3\sqrt{30}$
$=\frac{20\sqrt{3}}{3}-3\sqrt{30}$
$=\frac{20}{3}\sqrt{3} - 3\sqrt{30}$
(3) $4\sqrt{\frac{9}{8}}×\frac{1}{2}\sqrt{\frac{49}{50}}-\sqrt{\frac{9}{28}}÷\sqrt{1\frac{1}{35}}$
$=4×\frac{3}{2\sqrt{2}}×\frac{1}{2}×\frac{7}{5\sqrt{2}}-\sqrt{\frac{9}{28}÷\frac{36}{35}}$
$=4×\frac{3}{2\sqrt{2}}×\frac{7}{10\sqrt{2}}-\sqrt{\frac{9}{28}×\frac{35}{36}}$
$=\frac{84}{40}-\sqrt{\frac{5}{16}}$
$=\frac{21}{10}-\frac{\sqrt{5}}{4}$
(4) $\sqrt{45}÷2\sqrt{\frac{1}{5}}×\sqrt{2\frac{2}{3}}-2\sqrt{3}×(\frac{3}{2}\sqrt{2}-3\sqrt{\frac{3}{8}})$
$=3\sqrt{5}÷\frac{2}{\sqrt{5}}×\frac{2\sqrt{6}}{3}-2\sqrt{3}×(\frac{3}{2}\sqrt{2}-\frac{3\sqrt{6}}{4})$
$=3\sqrt{5}×\frac{\sqrt{5}}{2}×\frac{2\sqrt{6}}{3}-3\sqrt{6}+\frac{3\sqrt{18}}{2}$
$=5\sqrt{6}-3\sqrt{6}+\frac{9\sqrt{2}}{2}$
$=2\sqrt{6}+\frac{9}{2}\sqrt{2}$
(5) $(x+\sqrt{y})(x-\sqrt{y})-(2x-\sqrt{y})(x + 2\sqrt{y})$
$=x^{2}-y-(2x^{2}+4x\sqrt{y}-x\sqrt{y}-2y)$
$=x^{2}-y-2x^{2}-3x\sqrt{y}+2y$
$=-x^{2}+y - 3x\sqrt{y}$
(6) $(1 - 2\sqrt{3})(1 + 2\sqrt{3})-(2\sqrt{3}-1)^{2}$
$=1-(2\sqrt{3})^{2}-[(2\sqrt{3})^{2}-4\sqrt{3}+1]$
$=1 - 12 - (12 - 4\sqrt{3}+1)$
$=-11 - 13 + 4\sqrt{3}$
$=-24 + 4\sqrt{3}$
6. 比较$\sqrt{5}+\sqrt{2}$与$\sqrt{3}+2$的大小关系，并写出解答过程。
答案
6. 解: $ \because (\sqrt{5} + \sqrt{2})^{2} = 7 + 2\sqrt{10} $, $ (\sqrt{3} + 2)^{2} = 7 + 4\sqrt{3} $, $ 2\sqrt{10} = \sqrt{40} $, $ \sqrt{48} = 4\sqrt{3} $, $ \therefore 2\sqrt{10} ＜ 4\sqrt{3} $, $ \therefore \sqrt{5} + \sqrt{2} ＜ \sqrt{3} + 2 $.
解析
解: $\because (\sqrt{5} + \sqrt{2})^{2} = 5 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10}$,
$(\sqrt{3} + 2)^{2} = 3 + 4\sqrt{3} + 4 = 7 + 4\sqrt{3}$,
又$2\sqrt{10} = \sqrt{4 × 10} = \sqrt{40}$, $4\sqrt{3} = \sqrt{16 × 3} = \sqrt{48}$,
$\because \sqrt{40} ＜ \sqrt{48}$,
$\therefore 2\sqrt{10} ＜ 4\sqrt{3}$,
$\therefore 7 + 2\sqrt{10} ＜ 7 + 4\sqrt{3}$,
$\therefore (\sqrt{5} + \sqrt{2})^{2} ＜ (\sqrt{3} + 2)^{2}$,
$\because \sqrt{5} + \sqrt{2} ＞ 0$, $\sqrt{3} + 2 ＞ 0$,
$\therefore \sqrt{5} + \sqrt{2} ＜ \sqrt{3} + 2$.
$(\sqrt{3} + 2)^{2} = 3 + 4\sqrt{3} + 4 = 7 + 4\sqrt{3}$,
又$2\sqrt{10} = \sqrt{4 × 10} = \sqrt{40}$, $4\sqrt{3} = \sqrt{16 × 3} = \sqrt{48}$,
$\because \sqrt{40} ＜ \sqrt{48}$,
$\therefore 2\sqrt{10} ＜ 4\sqrt{3}$,
$\therefore 7 + 2\sqrt{10} ＜ 7 + 4\sqrt{3}$,
$\therefore (\sqrt{5} + \sqrt{2})^{2} ＜ (\sqrt{3} + 2)^{2}$,
$\because \sqrt{5} + \sqrt{2} ＞ 0$, $\sqrt{3} + 2 ＞ 0$,
$\therefore \sqrt{5} + \sqrt{2} ＜ \sqrt{3} + 2$.
