[Toán 8] Làm phép tính

T

thieukhang61

\[\begin{array}{l}
\frac{{3x + 1}}{{{{(x - 1)}^2}}} - \frac{1}{{x + 1}} + \frac{{x + 3}}{{1 - {x^2}}}\\
= \frac{{(3x + 1)(x + 1)}}{{{{(x - 1)}^2}(x + 1)}} - \frac{{{{(x - 1)}^2}}}{{{{(x - 1)}^2}(x + 1)}} + \frac{{(x + 3)(1 - x)}}{{{{(x - 1)}^2}(x + 1)}}\\
= \frac{{(3x + 1)(x + 1) - {{(x - 1)}^2} + (x + 3)(1 - x)}}{{{{(x - 1)}^2}(x + 1)}}\\
= \frac{{3{x^2} + 4x + 1 - {x^2} + 2x - 1 - {x^2} - 2x + 3}}{{{{(x - 1)}^2}(x + 1)}} = \frac{{{x^2} + 4x + 3}}{{{{(x - 1)}^2}(x + 1)}} = \frac{{(x + 1)(x + 3)}}{{{{(x - 1)}^2}(x + 1)}} = \frac{{x + 3}}{{{{(x - 1)}^2}}}
\end{array}\]
 
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