Đáp án:
$\frac{-32y+y^3}{(y+4)^2(y-4)^2}$
Giải thích các bước giải:
$C=\frac{1}{y^2+8y+16}+\frac{1}{8y-y^2-16}+\frac{y}{y^2-16}\\
=\frac{1}{y^2+2.4y+4^2}+\frac{1}{-(-2.4y+y^2+4^2)}+\frac{y}{(y-4)(y+4)}\\
=\frac{1}{(y+4)^2}+\frac{-1}{(y-4)^2}+\frac{y}{(y-4)(y+4)}\\
=\frac{(y-4)^2}{(y+4)^2(y-4)^2}+\frac{-(y+4)^2}{(y+4)^2(y-4)^2}+\frac{y(y^2-16)}{(y-4)^2(y+4)^2}\\
=\frac{(y-4)^2-(y+4)^2+y(y^2-16)}{(y+4)^2(y-4)^2}\\
=\frac{(y-4+y+4)(y-4-y-4)+y^3-16y}{(y+4)^2(y-4)^2}\\
=\frac{2y.(-8)+y^3-16y}{(y+4)^2(y-4)^2}\\
=\frac{-16y+y^3-16y}{(y+4)^2(y-4)^2}\\
=\frac{-32y+y^3}{(y+4)^2(y-4)^2}\\$
