40. Five mark exercise problem X

If \(A\) \(=\) \(\frac{x}{x + 1}\) and \(B\) \(=\) \(\frac{1}{x + 1}\), then prove that \(\frac{(A + B)^2 + (A - B)^2}{A \div B}\) \(=\) \(\frac{2(x^2 + 1)}{x (x + 1)^2}\).

 

Answer:

 

\((A + B)^2\) \(=\)

 

\((A - B)^2\) \(=\)

 

\(A \div B\) \(=\)

 

\(\frac{(A + B)^2 + (A - B)^2}{A \div B}\)  \(\frac{2(x^2 + 1)}{x (x + 1)^2}\)