Toán Lớp 8: P =($\dfrac{2}{x + 4}$ + $\dfrac{x + 20}{x² – 16}$) . $\dfrac{x – 4}{x + 5}$
=>Rút gọn P
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Toán Lớp 8: P =($\dfrac{2}{x + 4}$ + $\dfrac{x + 20}{x² – 16}$) . $\dfrac{x – 4}{x + 5}$
=>Rút gọn P
Comments ( 2 )
= ($\dfrac{2}{x+4}$ + $\dfrac{x+20}{(x-4)(x+4)}$) . $\dfrac{x-4}{x+5}$
= ($\dfrac{2(x-4)}{(x+4)(x-4)}$+$\dfrac{x+20}{(x-4)(x+4)}$) . $\dfrac{x-4}{x+5}$
= $\dfrac{2x -8 +x +20}{(x-4)(x+4)}$ . $\dfrac{x-4}{x+5}$
= $\dfrac{3x + 12}{(x-4)(x+4)}$ . $\dfrac{x-4}{x+5}$
= $\dfrac{3(x+4)}{(x-4)(x+4)}$ . $\dfrac{x-4}{x+5}$
= $\dfrac{3(x+4)(x-4)}{(x+4)(x-4)(x+5)}$
= $\dfrac{3}{x+5}$
P = \left(\dfrac{2}{x+4}+ \dfrac{x+20}{x^2 – 16}\right)\cdot \dfrac{x-4}{x+5}\qquad (x\ne \pm 4;x\ne -5)\\
\to P =\left[\dfrac{2(x-4)}{(x-4)(x+4)} + \dfrac{x+20}{(x-4)(x+4)}\right]\cdot \dfrac{x-4}{x+5}\\
\to P =\dfrac{2x- 8 + x + 20}{(x-4)(x+4)}\cdot \dfrac{x-4}{x+5}\\
\to P = \dfrac{3x + 12}{x+4}\cdot \dfrac{1}{x+5}\\
\to P =\dfrac{3(x+4)}{(x+4)(x+5)}\\
\to P = \dfrac{3}{x+5}
\end{array}\)