Question

38 If \( A=\frac{x}{x+1} B=\frac{1}{x+1} \) prove that \( \frac{(A+B) 2+(A-B) 2}{A \div B}=\frac{2(x 2+1)}{x(x+1) 2} \)

Answer 1

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.