यदि \(x=\sqrt{8}+3\) है तो \(x+\frac{1}{x-3}\) का मान क्या है?
If \(x=\sqrt{8}+3\), what is the value of \(x+\frac{1}{x-3}\)?
Explanation opens after your attempt
Correct Answer
A. \(3+\frac{5\sqrt{2}}{2}\)
Concept
\(x-3=\sqrt{8}=2\sqrt{2}\), so \( \frac{1}{x-3}=\frac{\sqrt{2}}{4} \). The total should be \(3+2\sqrt{2}+\frac{\sqrt{2}}{4}=3+\frac{9\sqrt{2}}{4}\).
Why this answer is correct
The correct answer is A. \(3+\frac{5\sqrt{2}}{2}\). \(x-3=\sqrt{8}=2\sqrt{2}\), so \( \frac{1}{x-3}=\frac{\sqrt{2}}{4} \). The total should be \(3+2\sqrt{2}+\frac{\sqrt{2}}{4}=3+\frac{9\sqrt{2}}{4}\).
Exam Tip
\(x-3=\sqrt{8}=2\sqrt{2}\) इसलिए \( \frac{1}{x-3}=\frac{\sqrt{2}}{4} \)। कुल \(3+2\sqrt{2}+\frac{\sqrt{2}}{4}=3+\frac{9\sqrt{2}}{4}\) होना चाहिए।
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