بسط الجمل العددية الآتية:

$\frac{\sqrt{12}}{3\sqrt[3]{125}}÷\frac{5\sqrt[3]{8}}{\sqrt{25}}$

$\frac{\sqrt{12}}{3\sqrt[3]{125}}÷\frac{5\sqrt[3]{8}}{\sqrt{25}}=\frac{2\sqrt{3}}{3\times 5}÷\frac{5\times 2}{5}=\frac{2\sqrt{3}}{15}÷\frac{10}{5}=\frac{2\sqrt{3}}{15}\times \frac{5}{10}=\frac{\sqrt{3}}{15}$

تدرب وحل التمرينات

(1)- بسط الجمل العددية الآتية:

$(\sqrt{18}-\sqrt{50}){\left(\frac{-27}{64}\right)}^{\frac{1}{3}}$

$(\sqrt{18}-\sqrt{50}){\left(\frac{-27}{64}\right)}^{\frac{1}{3}}=(3\sqrt{2}-5\sqrt{2}{)}^3\sqrt{\frac{-27}{64}}=(-2\sqrt{2})\times \left(\frac{-3}{4}\right)=\frac{6\sqrt{2}}{4}=\frac{3\sqrt{2}}{2}$

$\frac{\sqrt{12}}{3\sqrt[3]{125}}÷\frac{5\sqrt[3]{8}}{\sqrt{25}}$

$\frac{\sqrt{12}}{3\sqrt[3]{125}}÷\frac{5\sqrt[3]{8}}{\sqrt{25}}=\frac{2\sqrt{3}}{3\times 5}÷\frac{5\times 2}{5}=\frac{2\sqrt{3}}{15}÷\frac{10}{5}=\frac{2\sqrt{3}}{15}\times \frac{5}{10}=\frac{\sqrt{3}}{15}$

(2)- بسط الجملة العددية التالية واكتب الناتج لأقرب عشر:

$7\sqrt{\frac{2}{49}}-3\sqrt{\frac{8}{81}}+\sqrt{\frac{18}{36}}$

$\begin{array}{rl}& 7\sqrt{\frac{2}{49}}-3\sqrt{\frac{8}{81}}+\sqrt{\frac{18}{36}}=7\times \frac{\sqrt{2}}{7}-3\times \frac{\sqrt{8}}{9}+\frac{\sqrt{18}}{6}=\sqrt{2}-\frac{2\sqrt{2}}{3}+\frac{3\sqrt{2}}{6}\\ & =\frac{6\sqrt{2}-4\sqrt{2}+3\sqrt{2}}{6}=\frac{5\sqrt{2}}{6}=\frac{5\times 1.41}{6}=\frac{7.05}{6}=1.17\approx 1.2\end{array}$

(3)- بسط الجمل العددية التالية باستعمال تنسيب المقام وترتيب العمليات على الأعداد:

$\frac{\sqrt{7}-3\sqrt{5}}{\sqrt{7}+3\sqrt{5}}$

$\begin{array}{rl}& \frac{\sqrt{7}-3\sqrt{5}}{\sqrt{7}+3\sqrt{5}}=\frac{\sqrt{7}-3\sqrt{5}}{\sqrt{7}+3\sqrt{5}}\times \frac{\sqrt{7}-3\sqrt{5}}{\sqrt{7}-3\sqrt{5}}=\frac{(\sqrt{7}-3\sqrt{5})(\sqrt{7}-3\sqrt{5})}{(\sqrt{7}{)}^2-(3\sqrt{5}{)}^2}\\ =& \frac{\sqrt{7}\times \sqrt{7}-\sqrt{7}\times 3\sqrt{5}-3\sqrt{5}\times \sqrt{7}+3\sqrt{5}\times 3\sqrt{5}}{7-45}=\frac{7-3\sqrt{35}-3\sqrt{35}+45}{-38}=\frac{52-6\sqrt{35}}{-38}=\frac{-52+6\sqrt{35}}{38}\end{array}$

$\frac{\sqrt{33}-\sqrt{11}}{\sqrt{99}}-\frac{\sqrt{60}-\sqrt{5}}{5\sqrt{15}}$

$\begin{array}{rl}& \frac{\sqrt{33}-\sqrt{11}}{\sqrt{99}}-\frac{\sqrt{60}-\sqrt{5}}{5\sqrt{15}}=\frac{\sqrt{33}-\sqrt{11}}{3\sqrt{11}}-\frac{2\sqrt{15}-\sqrt{5}}{5\sqrt{15}}\\ =& \frac{\sqrt{33}-\sqrt{11}}{3\sqrt{11}}\times \frac{\sqrt{11}}{\sqrt{11}}-\frac{2\sqrt{15}-\sqrt{5}}{5\sqrt{15}}\times \frac{\sqrt{15}}{\sqrt{15}}=\frac{\sqrt{33}\times \sqrt{11}-\sqrt{11}\times \sqrt{11}}{3\sqrt{11}\times \sqrt{11}}-\frac{2\sqrt{15}\times \sqrt{15}-\sqrt{5}\times \sqrt{15}}{5\sqrt{15}\times \sqrt{15}}\\ =& \frac{\sqrt{3}\times \sqrt{11}\times \sqrt{11}-11}{3\times 11}-\frac{30-\sqrt{5}\times \sqrt{5}\times \sqrt{3}}{5\times 15}=\frac{11\sqrt{3}-11}{33}-\frac{30-5\sqrt{3}}{75}=\frac{11(\sqrt{3}-1)}{33}-\frac{5(6-\sqrt{3})}{75}\\ =& \frac{(\sqrt{3}-1)}{3}-\frac{(6-\sqrt{3})}{15}=\frac{5(\sqrt{3}-1)-(6-\sqrt{3})}{15}=\frac{5\sqrt{3}-5-6+\sqrt{3}}{15}=\frac{6\sqrt{3}-11}{15}\end{array}$