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Question Number 140534 by SOMEDAVONG last updated on 09/May/21

Answered by EDWIN88 last updated on 09/May/21

$$\left(\mathrm{i}\right)=\frac{16\mathrm{x}-24}{(\mathrm{x}-1)(\mathrm{x}-3)(\mathrm{x}+3)}=\frac{\mathrm{a}}{\mathrm{x}-1}+\frac{\mathrm{b}}{\mathrm{x}-3}+\frac{\mathrm{c}}{\mathrm{x}+3}$$$$\mathrm{a}={\left[\frac{16\mathrm{x}-24}{(\mathrm{x}-3)(\mathrm{x}+3)}\right]}_{\mathrm{x}=1}=\frac{-8}{-8}=1$$$$\mathrm{b}={\left[\frac{16\mathrm{x}-24}{(\mathrm{x}-1)(\mathrm{x}+3)}\right]}_{\mathrm{x}=3}=\frac{48-24}{6.2}=2$$$$\mathrm{c}={\left[\frac{16\mathrm{x}-24}{(\mathrm{x}-1)(\mathrm{x}-3)}\right]}_{\mathrm{x}=-3}=\frac{-72}{-4.(-6)}=-3$$$$\frac{16\mathrm{x}-24}{(\mathrm{x}-1)({\mathrm{x}}^2-9)}=\frac{1}{\mathrm{x}-1}+\frac{2}{\mathrm{x}-3}-\frac{3}{\mathrm{x}+3}$$

Commented by SOMEDAVONG last updated on 09/May/21

$$$$$$\mathrm{Thanks}\mathrm{so}\mathrm{much}$$

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