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What is the simplified form of (\sqrt{242}-\sqrt{128}+\sqrt{98}-\sqrt{72})?(\sqrt{242}-\sqrt{128}+\sqrt{98}-\sqrt{72}) का सरल रूप क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (4\sqrt{2})(4\sqrt{2})
We have (\sqrt{242}=11\sqrt{2}), (\sqrt{128}=8\sqrt{2}), (\sqrt{98}=7\sqrt{2}), and (\sqrt{72}=6\sqrt{2}). The total is (4\sqrt{2}).(\sqrt{242}=11\sqrt{2}), (\sqrt{128}=8\sqrt{2}), (\sqrt{98}=7\sqrt{2}), और (\sqrt{72}=6\sqrt{2})। कुल (4\sqrt{2}) मिलता है।
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