Question

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Номер 36
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100 баллов

Answer 1

$f(a)=\frac{1}{a-1}+2=\frac{1+2a-2}{a-1}=\frac{2a-1}{a-1}\\\\g(b)=b+\frac{1}{b} =\frac{b^{2}+1 }{b}\\\\f(a)-2g(b)=\frac{2a-1}{a-1} -\frac{2b^{2}+2 }{b} =\frac{2ab-b-2ab^{2}-2a+2b^{2}+2}{b(a-1)}$

$f(2a) = \frac{1}{2a-1} +2=\frac{1+4a-2}{a-1}=\frac{4a-1}{a-1}\\\\g(b)=b+\frac{1}{b}=\frac{b^{2}+1 }{b} \\\\-5f(2a)-\frac{2}{g(b)}=-5*\frac{4a-1}{a-1} -\frac{2}{\frac{b^{2}+1 }{b} }=\frac{-20a+5}{a-1}-\frac{2b}{b^{2}+1 }=\frac{-20ab^{2} -20a+5b^{2}+5-2ab+2b }{(a-1)(b^{2}+1) }$