Question

Найдите значения: 3 * f ( 1 / b ) / f * ( b ) если f ( b ) равно ( 9 / b + b ) * ( 1 / b + 9 b )
Помогите времени мало

Answer 1

$\displaystyle\bf\\f(b)=\Big(\frac{9}{b} +b\Big)\cdot\Big(\frac{1}{b} +9b\Big)=\frac{9+b^{2} }{b} \cdot\frac{1+9b^{2} }{b} =\frac{(9+b^{2})(1+9b^{2} ) }{b^{2} } \\\\\\f\Big(\frac{1}{b} \Big)=\frac{\Big[9+\Big(\dfrac{1}{b}\Big)^{2} \Big]\cdot\Big[1+9\cdot\Big(\dfrac{1}{b} \Big)^{2}\Big] }{\Big(\dfrac{1}{b} \Big)^{2} } =\frac{\Big(9+\dfrac{1}{b^{2} } \Big)\Big(1+\dfrac{9}{b^{2} }\Big) }{\dfrac{1}{b^{2} } } =$

$\displaystyle\bf\\=b^{2} \cdot\frac{9b^{2} +1}{b^{2} } \cdot\frac{b^{2}+9 }{b^{2} }=\frac{(9b^{2} +1)\cdot(b^{2}+9) }{b^{2} } \\\\\\3\cdot f\Big(\frac{1}{b} \Big):f\Big(b\Big)=3\cdot\frac{(9b^{2} +1)\cdot(b^{2}+9) }{b^{2} } :\frac{(1+9b^{2} )\cdot(9+b^{2}) }{b^{2} } =\\\\\\=3\cdot \frac{(9b^{2} +1)\cdot(b^{2}+9) }{b^{2} } \cdot\frac{b^{2} }{(1+9b^{2})(9+b^{2} ) } =\boxed3$