Question
Download Solution PDFयदि \(7 b-\frac{1}{4 b}=7\) है, तो \(16 b^2+\frac{1}{49 b^2}\) का मान क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFप्रयुक्त सूत्र
(a - b)2 = a2 + b2 - 2ab
गणना
व्यंजक को 4/7 से गुणा करने पर
⇒ 4/7 × (7b - 1/4b) = 7 × 4/7
⇒ 4b - 1/7b = 4
दोनों पक्षों का वर्ग करने पर:
⇒ (4b - 1/7b)2 = 42
⇒ \(16 b^2+\frac{1}{49 b^2}\)- 2 × 4 × 1/7 = 16
⇒ \(16 b^2+\frac{1}{49 b^2}\) = 16 + 8/7
⇒ \(16 b^2+\frac{1}{49 b^2}\) = 120/7
मान 120/7 है।
Last updated on Jun 13, 2025
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