UPSKILL MATH PLUS
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\(=\)
\(1\)
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:
\((A + B)^2\) \(=\)
\((A - B)^2\) \(=\)
\(A \div B\) \(=\)
\(\frac{(A + B)^2 + (A - B)^2}{A \div B}\) \(\frac{2(x^2 + 1)}{x (x + 1)^2}\)
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