यदि \(x=1+\sqrt{2}\), तो \(x+\frac{1}{x}\) का मान क्या है?

If \(x=1+\sqrt{2}\), what is \(x+\frac{1}{x}\)?

Explanation opens after your attempt
Correct Answer

A. \(2\sqrt{2}\)

Step 1

Concept

\(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\), so the sum is \(2\sqrt{2}\). Rationalising the denominator is a quick exam method.

Step 2

Why this answer is correct

The correct answer is A. \(2\sqrt{2}\). \(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\), so the sum is \(2\sqrt{2}\). Rationalising the denominator is a quick exam method.

Step 3

Exam Tip

\(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\), इसलिए योग \(2\sqrt{2}\) है। परीक्षा में हर का परिमेयकरण तेज तरीका है।

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Mathematics Answer, Explanation and Revision Hints

यदि \(x=1+\sqrt{2}\), तो \(x+\frac{1}{x}\) का मान क्या है? / If \(x=1+\sqrt{2}\), what is \(x+\frac{1}{x}\)?

Correct Answer: A. \(2\sqrt{2}\). Explanation: \(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\), इसलिए योग \(2\sqrt{2}\) है। परीक्षा में हर का परिमेयकरण तेज तरीका है। / \(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\), so the sum is \(2\sqrt{2}\). Rationalising the denominator is a quick exam method.

Which concept should I revise for this Mathematics MCQ?

\(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\), so the sum is \(2\sqrt{2}\). Rationalising the denominator is a quick exam method.

What exam hint can help solve this Mathematics question?

\(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\), इसलिए योग \(2\sqrt{2}\) है। परीक्षा में हर का परिमेयकरण तेज तरीका है।