3 задание с решением пж

A) $x= \frac{4}{x} - \frac{x}{9} +2$
   $x= \frac{4}{x} - \frac{x}{9} +2, x \neq 0$
   $x- \frac{4}{x} + \frac{x}{9} -2=0$
   $\frac{9x+x}{9} - \frac{4}{x} -2=0$
   $\frac{10x}{9} - \frac{4}{x} -2=0$
   $\frac{10x^{2}-36-18x }{9x}=0$
   $10x^{2}-36-18x =0$
   $2(5x^{2}-18-9x) =0$
   $2(5x^{2}-9x-18) =0$
   $2(5x^{2}+6x-15x-18) =0$
   $2(x(5x+6)-3(5x+6)) =0$
   $2(x-3)(5x+6) =0$
   $(x-3)(5x+6) =0$
   $\left \{ {{x-3} \atop {5x+6=0}} \right.$
   $\left \{ {{x=3} \atop {x=-1.2}} \right.$
б) $(1+ \frac{1}{x})^{2}= \frac{9}{ x^{2}}$
   $(\frac{x+1}{x})^{2}= \frac{9}{ x^{2}}$
   $(\frac{x+1}{x})^{2}= \frac{9}{ x^{2}}, x \neq 0$
   $\frac{(x+1)^{2}}{x^{2}}}= \frac{9}{ x^{2}}$
   $\frac{x^{2}+2x+1}{x^{2}}}= \frac{9}{ x^{2}}$
   $x^{2}+2x+1=9$
   $x^{2}+2x+1-9=0$
   $x^{2}+2x-8=0$
   $D=b^{2}-4ac=2^{2}-4*1*(-8)=4+32=36$
   $\left\{{{x=\frac{-2+6}{2} }\atop{x=\frac{-2-6}{2}}}\right.$
   $\left\{{{x=2}\atop{x=-4}}\right.$