0,\; t.k.\; D=-48<0\\\\\frac{1}{x+4}=\frac{2}{17}\\\\2x+8=17\\\\2x=9\\\\x=4,5" alt="\frac{x^2-4x+16}{x^3+64}=\frac{2}{17}\\\\\frac{x^2-4x+16}{(x+4)(x^2-4x+16)}=\frac{2}{17}\; ,\; \; x\ne 4\; ;\; x^2-4x+16>0,\; t.k.\; D=-48<0\\\\\frac{1}{x+4}=\frac{2}{17}\\\\2x+8=17\\\\2x=9\\\\x=4,5" align="absmiddle" class="latex-formula">