$\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\frac{x-1}{x+2}+\frac{x+3}{x-4}=\frac{2}{\left(x-2\right)\left(4-x\right)}\frac{7x-3}{x-x^3}=\frac{1}{x-1}-\frac{5}{\left(x-1\right)}\frac{1}{x+2}+\frac{1}{x+3}=\frac{1}{\left(x+2\right)\left(x+3\right)}\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\left(\frac{x^3-3x}{x^2-9}\right):\left(\frac{9-x^2}{\left(x+3\right).\left(x+2\right)}+\frac{x-3}{x-2}-\frac{x+2}{x+3}\right)\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{3}-\frac{5\left(0,4-2x\right)}{6}\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{\left(x-1\right)\left(x+3\right)}\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=-\frac{7}{6\left(x+5\right)}\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}$ \frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{3}-\frac{5\left(0,4-2x\right)}{6}$$ \iff \frac{21\left(9x-0,7\right)}{84}-\frac{12\left(5x-1,5\right)}{84}$$ \frac{28\left(7x-1,1\right)}{84}-\frac{70\left(0,4-2x\right)}{84}$$ \begin{array}{c}\iff 21(9\end{array}$\left(\frac{2x}{3}+4\right)\left(2x-3\right)\left(\frac{x}{2}-1\right)=0\left(x^2-4\right)\left(\frac{x}{4}-\frac{4x+5}{3}\right)=0\left(\frac{7}{x-2}+\frac{8}{x+2}\frac{14x}{x^2-4}\right):\left(\frac{1}{x-2}-\frac{1}{x+2}\right)\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0M=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\mathrm{b,\; N}=\frac{\mathrm{bc}}{a^2+2bc}+\frac{\mathrm{ac}}{b^2+2ac}+\frac{\mathrm{ab}}{c^2+2ab}\left(2n-3\right).n-2n.\left(n+2\right)⋮7\forall n\in Zn.\left(2n-3\right)-2n.\left(n+1\right)⋮5\forall n\in Z5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x\; +3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}{\left(x+\frac{1}{5}\right)}^2+\frac{17}{25}=\frac{26}{25}\frac{}{}$