Question

a²-9/2a² * a/2a+b при a=1/27
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Answer 1

$\frac{ a^{2} -9}{2a^{2} } * \frac{a}{2a+b} = \frac{(a-3)(a+3)}{2a(2a+b)}$

при  а=1/27

$\frac{( \frac{1}{27} -3)( \frac{1}{27} +3)}{2* \frac{1}{27} (2* \frac{1}{27} +b)}=- \frac{2 \frac{26}{27}*3 \frac{1}{27} }{ \frac{2}{27}( \frac{2}{27} +b) } =- \frac{80*82}{27*27} :( \frac{4}{27*27} + \frac{2b}{27} )= \\ \\ - \frac{80*82}{27*27} : \frac{4+54b}{27*27} =-\frac{80*82}{27*27}* \frac{27*27}{2(2+17b)} =- \frac{3280}{2+17b}$