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بسط الجمل العددية التالية باستعمال الخواص (الأبدال، التجميع، التوزيع):

$4\sqrt{3}-9\sqrt{3}$

$-5\sqrt{3}$

$\sqrt{7}(\sqrt{7}+\sqrt{2})$

$7+\sqrt{7\times 2}=7+\sqrt{14}$

$3\sqrt{12}-5\sqrt{8}$

$3\sqrt{4\times 3}-5\sqrt{4\times 2}=6\sqrt{3}-10\sqrt{2}$

$\sqrt{6}(2\sqrt{3}+6\sqrt{2})$

$\begin{array}{rl}& \sqrt{6}\times 2\sqrt{3}+\sqrt{6}\times 6\sqrt{2}\\ & =2\sqrt{3}\times \sqrt{3\times 2}+6\sqrt{2}\sqrt{2\times 3}\end{array}$

$\sqrt{27}(5-\sqrt{3})$

$3\sqrt{3}(5-\sqrt{3})=15\sqrt{3}-9$

$\sqrt{2}(\sqrt{6}-3\sqrt{2})$

$\sqrt{12}-12=2\sqrt{3}-12$

بسط الجمل العددية التالية باستعمال الخواص (العنصر المحايد، النظير الجمعي والنظير الضربي):

$\frac{\sqrt{7}-2}{\sqrt{3}}$

$$\begin{gathered} \frac{\sqrt{7}-2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ \begin{array}{rl}& =\frac{\sqrt{3}(\sqrt{7}-2)}{\sqrt{3}\sqrt{3}}=\frac{\sqrt{3}\sqrt{7}-2\sqrt{3}}{3}\\ & =\sqrt{21}-2\sqrt{3}\end{array} \end{gathered}$$

$\frac{8+\sqrt{7}}{4\sqrt{5}}$

$\frac{8+\sqrt{7}}{4\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{8\sqrt{5}+\sqrt{35}}{20}$

$\frac{\sqrt{8}-1}{2\sqrt{8}}$

$\frac{\sqrt{8}-1}{2\sqrt{8}}\times \frac{\sqrt{8}}{\sqrt{8}}=\frac{8-\sqrt{8}}{16}$

$\frac{\sqrt{12}-2}{2\sqrt{3}}$

$\frac{\sqrt{12}-2}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{3-\sqrt{3}}{2}$

$\frac{\sqrt{20}+\sqrt{5}}{\sqrt{5}}$

$\frac{2\sqrt{5}+\sqrt{5}}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}}=3$

$\frac{12-6\sqrt{3}}{\sqrt{27}}$

$\frac{12-6\sqrt{3}}{\sqrt{27}}\times \frac{\sqrt{27}}{\sqrt{27}}==\frac{4\sqrt{3}-6}{3}$

بسط الجمل العددية التالية باستعمال خصائص الأعداد الحقيقية:

$\sqrt{3}(5-\sqrt{3})$

$5\sqrt{3}-3$

$\sqrt{7}(\sqrt{7}+\sqrt{28})$

$7+\sqrt{28\times 7}=7+\sqrt{196}=21$

$4\sqrt{32}+3\sqrt{2}-5\sqrt{2}$

$16\sqrt{2}+3\sqrt{2}-5\sqrt{2}=14\sqrt{2}$

$\frac{1}{4}\sqrt{11}+\frac{1}{5}\sqrt{11}-\frac{1}{2}\sqrt{11}$

$(\frac{1}{4}+\frac{1}{5}-\frac{1}{2})\sqrt{11}=-\frac{\sqrt{11}}{20}$

$\frac{\sqrt{24}-5}{2\sqrt{3}}$

$\frac{\sqrt{24}-5}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{3\sqrt{7}-5\sqrt{3}}{6}$

$\frac{7\sqrt{3}}{\sqrt{2}}÷\frac{14\sqrt{3}}{\sqrt{6}}$

$\frac{7\sqrt{3}}{\sqrt{2}}\times \frac{\sqrt{2}\sqrt{3}}{14\sqrt{3}}=\frac{3}{2\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{3\sqrt{3}}{6}$