Question

Докажите тождество: (z/z-2 - z^2/z^3+8 * z^2+2z/z-2) : 8/z^2-2z+4 + z^2+z+6/4z+8 = 6-z/4z+8

Answer 1

$(\frac{z}{z-2}-\frac{z^2}{z^3+8}*\frac{z^2+2z}{z-2}):\frac{8}{z^2-2z+4}+\frac{z^2+z+6}{4z+8}$

$1) \frac{z^2}{z^3+8}*\frac{z^2+2z}{z-2}=\frac{z^2}{(z+2)(z^2-2z+4)}*\frac{z(z+2)}{z-2}=\frac{z^3}{(z^2-2z+4)(z-2)} \\ \\ 2) \frac{z}{z-2}-\frac{z^3}{(z^2-2z+4)(z-2)}=\frac{z(z^2-2z+4)-z^3}{(z^2-2z+4)(z-2)}=\frac{z^3-2z^2+4z-z^3}{(z^2-2z+4)(z-2)}= \\ \\ =\frac{-2z^2+4z}{(z^2-2z+4)(z-2)}=\frac{-2z(z-2)}{(z^2-2z+4)(z-2)}=\frac{-2z}{z^2-2z+4}$

$3) \frac{-2z}{z^2-2z+4}:\frac{8}{z^2-2z+4}=\frac{-2z}{z^2-2z+4}*\frac{z^2-2z+4}{8}=-\frac{z}{4} \\ \\ 4) -\frac{z}{4}+\frac{z^2+z+6}{4z+8}=\frac{z^2+z+6}{4(z+2)}-\frac{z}{4}= \frac{z^2+z+6-z(z+2)}{4(z+2)}=\frac{z^2+z+6-(z^2+2z)}{4(z+2)} = \\ \\ =\frac{z^2+z+6-z^2-2z}{4(z+2)}=\frac{6-z}{4(z+2)}=\frac{6-z}{4z+8}$

ответ: тождество верно