جواب تمرین صفحه ۷۶ ریاضی نهم

الف) ابتدا هر رادیکال را به صورت حاصل‌ضرب یک مربع کامل در یک عدد می‌نویسیم:

\[2\sqrt{50}=2\sqrt{5^2\times 2}=2\times 5\sqrt{2}=10\sqrt{2}\]
\[\sqrt{32}=\sqrt{4^2\times 2}=4\sqrt{2}\]
\[2\sqrt{72}=2\sqrt{6^2\times 2}=2\times 6\sqrt{2}=12\sqrt{2}\]
\[2\sqrt{50}+\sqrt{32}+2\sqrt{72}=10\sqrt{2}+4\sqrt{2}+12\sqrt{2}=26\sqrt{2}\]

ب)
\[\sqrt{8}=\sqrt{2^2\times 2}=2\sqrt{2}\]
\[\sqrt{128}=\sqrt{8^2\times 2}=8\sqrt{2}\]
\[\sqrt{50}=\sqrt{5^2\times 2}=5\sqrt{2}\]
\[\sqrt{8}+\sqrt{128}-\sqrt{50}=2\sqrt{2}+8\sqrt{2}-5\sqrt{2}=5\sqrt{2}\]

ج)
\[\sqrt[3]{27^2}=\sqrt[3]{(3^3)^2}=\sqrt[3]{3^6}=3^2=9\]

د)
\[\sqrt[3]{\frac{-27}{64}}=\frac{\sqrt[3]{-27}}{\sqrt[3]{64}}=\frac{-3}{4}\]

هـ)
\[(\sqrt{2}-\sqrt{5})(\sqrt{10}+\sqrt{2})=\sqrt{2}\sqrt{10}+\sqrt{2}\sqrt{2}-\sqrt{5}\sqrt{10}-\sqrt{5}\sqrt{2}\]
\[=\sqrt{20}+2-\sqrt{50}-\sqrt{10}\]
\[\sqrt{20}=2\sqrt{5},\qquad \sqrt{50}=5\sqrt{2}\]
\[=2\sqrt{5}+2-5\sqrt{2}-\sqrt{10}\]

و)
\[2\sqrt{48}=2\sqrt{4^2\times 3}=8\sqrt{3}\]
\[3\sqrt{27}=3\sqrt{3^2\times 3}=9\sqrt{3}\]
\[2\sqrt{48}-3\sqrt{27}=8\sqrt{3}-9\sqrt{3}=-\sqrt{3}\]