WAEC SSCE - General Mathematics - 1992…

Write as a single fraction $\frac{1}{1 - x} + \frac{2}{1 + x}$

$\frac{x + 3}{1 - x^2}$

$\frac{3 - x}{(1 - x)^2}$

$\frac{3 - x}{1 + x^2}$

$\frac{3 - x}{(1 + x)^2}$

$\frac{3 - x}{1 - x^2}$ Resposta correta

Detalhes da Resposta

To add the fractions $\frac{1}{1-x}$ and $\frac{2}{1+x}$, we need to have a common denominator. We can obtain this by multiplying the first fraction by $(1+x)/(1+x)$ and the second fraction by $(1-x)/(1-x)$. This gives us: $$ \frac{1}{1-x} \cdot \frac{1+x}{1+x} + \frac{2}{1+x} \cdot \frac{1-x}{1-x} = \frac{1+x}{1-x^2} + \frac{2(1-x)}{1-x^2} $$ Combining the two fractions gives us: $$ \frac{1+x+2(1-x)}{1-x^2} = \frac{3-x}{1-x^2} $$ Therefore, the single fraction is $\boxed{\frac{3-x}{1-x^2}}$.






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WAEC SSCE - General Mathematics - 1992 (Objective)