कौन सा विकल्प \(\frac{2}{\sqrt{7}-\sqrt{3}}\) का सरल रूप है?

Which option is the simplified form of \(\frac{2}{\sqrt{7}-\sqrt{3}}\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{\sqrt{7}+\sqrt{3}}{2}\)

Step 1

Concept

Multiplying by the conjugate makes the denominator (7-3=4). Hence we get (\frac{2\(\sqrt{7}+\sqrt{3}\)}{4}).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{\sqrt{7}+\sqrt{3}}{2}\). Multiplying by the conjugate makes the denominator (7-3=4). Hence we get (\frac{2\(\sqrt{7}+\sqrt{3}\)}{4}).

Step 3

Exam Tip

संयुग्मी से गुणा करने पर हर (7-3=4) बनता है। इसलिए (\frac{2\(\sqrt{7}+\sqrt{3}\)}{4}) मिलता है।

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Mathematics Answer, Explanation and Revision Hints

कौन सा विकल्प \(\frac{2}{\sqrt{7}-\sqrt{3}}\) का सरल रूप है? / Which option is the simplified form of \(\frac{2}{\sqrt{7}-\sqrt{3}}\)?

Correct Answer: A. \(\frac{\sqrt{7}+\sqrt{3}}{2}\). Explanation: संयुग्मी से गुणा करने पर हर (7-3=4) बनता है। इसलिए (\frac{2\(\sqrt{7}+\sqrt{3}\)}{4}) मिलता है। / Multiplying by the conjugate makes the denominator (7-3=4). Hence we get (\frac{2\(\sqrt{7}+\sqrt{3}\)}{4}).

Which concept should I revise for this Mathematics MCQ?

Multiplying by the conjugate makes the denominator (7-3=4). Hence we get (\frac{2\(\sqrt{7}+\sqrt{3}\)}{4}).

What exam hint can help solve this Mathematics question?

संयुग्मी से गुणा करने पर हर (7-3=4) बनता है। इसलिए (\frac{2\(\sqrt{7}+\sqrt{3}\)}{4}) मिलता है।