$$ 5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x\; +3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}$$ {\left(x+\frac{1}{5}\right)}^2+\frac{17}{25}=\frac{26}{25}$$ \frac{}{}$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0M=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\mathrm{b,\; N}=\frac{\mathrm{bc}}{a^2+2bc}+\frac{\mathrm{ac}}{b^2+2ac}+\frac{\mathrm{ab}}{c^2+2ab}$ \frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}$$ \frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}$$ \frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}$$ \frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}$$ \frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}$$ \frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}$\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\le \frac{a+b+c}{2abc}$ \left(\frac{2x}{3}+4\right)\left(2x-3\right)\left(\frac{x}{2}-1\right)=0$$ \left(x^2-4\right)\left(\frac{x}{4}-\frac{4x+5}{3}\right)=0$$ \frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}$$ \frac{x-1}{x+2}+\frac{x+3}{x-4}=\frac{2}{\left(x-2\right)\left(4-x\right)}$$ \frac{7x-3}{x-x^3}=\frac{1}{x-1}-\frac{5}{\left(x-1\right)}$$ \frac{1}{x+2}+\frac{1}{x+3}=\frac{1}{\left(x+2\right)\left(x+3\right)}$$ \left(\frac{x^3-3x}{x^2-9}\right):\left(\frac{9-x^2}{\left(x+3\right).\left(x+2\right)}+\frac{x-3}{x-2}-\frac{x+2}{x+3}\right)$$ P=5{\mathrm{x^}}^2+2{\mathrm{y^}}^2+4xy-4x+8y+25$$ \Delta DEM$$ \Delta ABC$$ \left(\frac{7}{x-2}+\frac{8}{x+2}\frac{14x}{x^2-4}\right):\left(\frac{1}{x-2}-\frac{1}{x+2}\right)$$