$$ \frac{}{}$$ \frac{\mathrm{15x}}{\mathrm{x2\; +\; 3x\; -4}}$$ \frac{}{}$\frac{AD}{DB}=\frac{CE}{EA}$ \frac{x}{2x+2}$$ \frac{2x}{x^2-2x-3}$$ \frac{x}{6-2x}$$ \frac{5}{-x^2+5x-6}$$ \frac{x+3}{2-x}$$ \frac{3}{x^2+x-2}$$ \frac{1}{x-1}$$ \frac{-4}{x+2}$$ \frac{x}{2x+2}$$ \frac{2x}{x^2-2x-3}$$ \frac{x}{6-2x}$$ \frac{5}{-x^2+5x-6}$$ \frac{x+3}{2-x}$$ \frac{3}{x^2+x-2}$$ \frac{1}{x-1}$$ \frac{-4}{x+2}$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0M=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\mathrm{b,\; N}=\frac{\mathrm{bc}}{a^2+2bc}+\frac{\mathrm{ac}}{b^2+2ac}+\frac{\mathrm{ab}}{c^2+2ab}$ \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}$$ \frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}$\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\le \frac{a+b+c}{2abc}$ \frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}$$ \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)}$$ x,y$$ {\mathrm{x2}}^{}={\mathrm{y2}}^{}+\backslash (\backslash sqrt\{y+1\}\backslash )$$ (x-y)(x+y)=\backslash (\backslash sqrt\{y+1\}\backslash )>0$$ x>y$$ x\ge 1$$ x+y\ge y+1\ge 1$$ x-y>0$$ x-y\ge 1$$ (x-y)(x+y)\ge y+1\ge \backslash (\backslash sqrt\{y+1\}\backslash )$$ "="$$ y+1=1;x+y=y+1;x-y=1$$ x=1;y=0$$