Question

помогите пожалуйста))

Answer 1

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