Question

Довести тотожність
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Answer 1

$\displaystyle \frac{1}{2z^2+5z}-\frac{2}{25-10z}-\frac{4}{4z^2-25}=\frac{1}{5z}\\\\\\ \frac{1}{z(2z+5)}+\frac{2}{5(2z-5)}-\frac{4}{(2z-5)(2z+5)}=\frac{1}{5z}\\\\\\\frac{5(2z-5)+2\cdot z(2z+5)-5z\cdot4}{5z(2z-5)(2z+5)}}=\frac{1}{5z}\\\\\\\frac{10z-25+4z^2+10z-20z}{5z(2z-5)(2z+5)}=\frac{1}{5z}\\\\\\\frac{4z^2-25}{5z(2z-5)(2z+5)}=\frac{1}{5z}\\\\\\\frac{(2z-5)(2z+5)}{5z(2z-5)(2z+5)}=\frac{1}{5z}\\\\\\\frac{1}{5z}\equiv\frac{1}{5z}$