Question

Пожалуйста, решите подробно.​

Answer 1

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

Ответ: 12/b.