\(\sqrt{242}-\sqrt{128}+\sqrt{72}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{242}-\sqrt{128}+\sqrt{72}\)?
Explanation opens after your attempt
Correct Answer
C. \(9\sqrt{2}\)
Concept
\(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). Therefore the result is \(9\sqrt{2}\).
Why this answer is correct
The correct answer is C. \(9\sqrt{2}\). \(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). Therefore the result is \(9\sqrt{2}\).
Exam Tip
\(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\) और \(\sqrt{72}=6\sqrt{2}\) है। इसलिए परिणाम \(9\sqrt{2}\) है।
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