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38 If \( A=\frac{x}{x+1} B=\frac{1}{x+1} \) prove that \( \frac{(A+B) 2+(A-B) 2}{A \div B}=\frac{2(x 2+1)}{x(x+1) 2} \) |
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Answer» We have, A = \(\frac{x}{x+1}\), B = \(\frac{1}{x+1}\) \(\frac{(A+B)^2+(A-B)^2}{A\,÷\,B}\) = \(\frac{B[(A^2+B^2+2AB)+(A^2+B^2-2AB)]}{A}\) = \(\frac{2B(A^2+B^2)}{A}\) = \(\frac{\frac{2}{x+1}((\frac{x}{x+1})^2+(\frac{1}{x+1})^2)}{\frac{x}{x+1}}\) = \(\frac{2}{x}\)\((\frac{x^2+1}{(x+1)^2})\) = \(\frac{2(x^2+1)}{x(x+1)^2}\) Hence proved. |
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