यदि \(x=\frac{1}{\sqrt{2}+1}\) है, तो \(x+\sqrt{2}\) का मान क्या है?
If \(x=\frac{1}{\sqrt{2}+1}\), what is the value of \(x+\sqrt{2}\)?
Explanation opens after your attempt
Correct Answer
A. (1)
Concept
\(\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\). So (x+\sqrt{2}=\(\sqrt{2}-1\)+\sqrt{2}=2\sqrt{2}-1).
Why this answer is correct
The correct answer is A. (1). \(\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\). So (x+\sqrt{2}=\(\sqrt{2}-1\)+\sqrt{2}=2\sqrt{2}-1).
Exam Tip
\(\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\) होता है। इसलिए \(x+\sqrt{2}=2\sqrt{2}-1\) नहीं, बल्कि (\(\sqrt{2}-1\)+\sqrt{2}=2\sqrt{2}-1) है।
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