Hard Mathematics Polynomials Class 10 Level 25

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

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

Explanation opens after your attempt
Correct Answer

A. \(5+2\sqrt{6}\)

Step 1

Concept

Multiplying by the conjugate makes the denominator (1). The numerator is (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}).

Step 2

Why this answer is correct

The correct answer is A. \(5+2\sqrt{6}\). Multiplying by the conjugate makes the denominator (1). The numerator is (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}).

Step 3

Exam Tip

हर के संयुग्मी से गुणा करने पर हर (1) बनता है। अंश (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}) है।

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Mathematics Answer, Explanation and Revision Hints

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

Correct Answer: A. \(5+2\sqrt{6}\). Explanation: हर के संयुग्मी से गुणा करने पर हर (1) बनता है। अंश (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}) है। / Multiplying by the conjugate makes the denominator (1). The numerator is (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}).

Which concept should I revise for this Mathematics MCQ?

Multiplying by the conjugate makes the denominator (1). The numerator is (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}).

What exam hint can help solve this Mathematics question?

हर के संयुग्मी से गुणा करने पर हर (1) बनता है। अंश (\(\sqrt{3}+\sqrt{2}\)2=5+2\sqrt{6}) है।

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