Toán Lớp 8: B = ($\frac{2x+1}{x-1}$+ $\frac{8}{x^{2}-1}$- $\frac{x-1}{x+1}$). $\frac{x^{2}-1}{5}$
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Toán Lớp 8: B = ($\frac{2x+1}{x-1}$+ $\frac{8}{x^{2}-1}$- $\frac{x-1}{x+1}$). $\frac{x^{2}-1}{5}$
Comments ( 2 )
DK:x \ne \pm 1\\
B = \left( {\dfrac{{2x + 1}}{{x – 1}} + \dfrac{8}{{{x^2} – 1}} – \dfrac{{x – 1}}{{x + 1}}} \right).\dfrac{{{x^2} – 1}}{5}\\
= \dfrac{{\left( {2x + 1} \right)\left( {x + 1} \right) + 8 – {{\left( {x – 1} \right)}^2}}}{{{x^2} – 1}}.\dfrac{{{x^2} – 1}}{5}\\
= \dfrac{{2{x^2} + 3x + 1 + 8 – {x^2} + 2x – 1}}{5}\\
= \dfrac{{{x^2} + 5x + 8}}{5}
\end{array}\)