Помогите пожалуйста))

Решите задачу:

$9.\;\left(\frac{b^{0,5}+3}{b^{1,5}-3b}-\frac{b^{0,5}-3}{b^{1,5}+3b}\right)\cdot\frac{b^{-9}}{b^{0,5}}=\left(\frac{(b^{0,5}+3)(b^{1,5}+3b)-(b^{0,5}-3)(b^{1,5}-3b)}{(b^{1,5}-3b)(b^{1,5}+3b)}\right)\cdot\frac{b^{-9}}{b^{0,5}}=\\=\frac{b^{0,5}\cdot b^{1,5}+3b\cdot b^{0,5}+3b^{1,5}+9b-b^{0,5}\cdot b^{1,5}+3b\cdot b^{0,5}+3b^{1,5}-9b}{b^3-9b^2}\cdot\frac{b^{-9}}{b^{0,5}}=$
$=\frac{6b^{1+0,5}+6b^{1,5}}{b^3-9b^2}\cdot\frac{1}{b^{9}\cdot b^{0,5}}=\frac{6b^{1,5}+6b^{1,5}}{b^3-9b^2}\cdot\frac{1}{b^{9,5}}=\frac{12b^{1,5}\cdot b^{-9,5}}{b^3-9b^2}}=\frac{12b^{-8}}{b^3-9b^2}=\\=\frac{12}{b^8(b^3-9b^2)}=\frac{12}{b^{11}-9b^{10}}$
$10.\;\sqrt{9x^2-6x+1}=x^2+1\\\sqrt{(3x-1)^2}=x^2+1\\\begin{cases}3x-1=x^2+1\\1-3x=x^2+1\end{cases}\Rightarrow\begin{cases}x^2-3x+2=0\\x^2+3x=0\end{cases}\\x^2-3x+2=0\\D=9-4\cdot2=1\\x_{1,2}=\frac{3\pm1}2\\x_1=2,\;x_2=1\\x^2+3x=0\\x(x+3)=0\\x_3=0,\;x_4=-3$