Question

Пожалуйста, решите подробно.​

Answer 1

$(\frac{b^{0.5}+3}{b^{1.5}-3b}^{(b^{1.5}+3b} -\frac{b^{0.5}-3}{b^{1.5}+3b}^{(b^{1.5}-3b} )*\frac{b-9}{b^{0.5}} =\\\frac{(b^{0.5}+3)(b^{1.5}+3b)-(b^{0.5}-3)(b^{1.5}-3b)}{(b^{1.5})^2-9b^2} *\frac{b-9}{b^{0.5}} =\frac{\mathfrak{b^2}+3b^{1.5}+3b^{1.5}+\mathfrak{9b}-(\mathfrak{b^2}-3b^{1.5}-3b^{1.5}+\mathfrak{9b})}{b^3-9b^2} *\frac{b-9}{b^{0.5}}=\\\frac{12b^{1.5}}{b^2\mathfrak{(b-9)}} *\frac{\mathfrak{b-9}}{b^{0.5}} =\\\frac{12b*\mathfrak{b^{0.5}}}{b^2*\mathfrak{b^{0.5}}} =\frac{12}{b}$

Ответ: 12/b.