Question

$\frac{9x^{2}+4 }{3x} +\frac{3x}{9x^{2} +9x+4} +5.5=0$

Answer 1

одз

$\left \{ {{x}\neq0\atop {9x^2+9x+4}\neq 0} \right. ;D<0;x\neq 0$

$\frac{9x^{2}+4 }{3x} +\frac{3x}{9x^{2} +9x+4} +5.5=0\\ \\\frac{9x^{2}+4 }{3x} +\frac{9x}{3x} -\frac{9x}{3x}+\frac{3x}{9x^{2} +9x+4} +5.5=0\\ \\\frac{9x^{2}+9x+4 }{3x} +\frac{3x}{9x^{2} +9x+4} +2.5=0\\ \\t=\frac{9x^{2}+9x+4 }{3x}\\ \\t+ \frac{1}{t} +2.5=0\\ \\t^2+2.5t+1=0 \\ \\D=6.25-4=2.25=1.5^2$

нет корней

$t_{2} =(-2.5-1.5)/2=-2 \\ \\ \frac{9x^2+9x+4}{3x} =-2\\ \\9x^2+15x+4=0 \\ \\D=225-4*9*4=224-144=81=9^2 \\ \\x_{1} =(-15-9)/18=-\frac{4}{3} \\ \\x_{2} =(-15+9)/18=-\frac{1}{3}$