$\frac{\frac{\mathrm{câu\; 2:\; 2x^2-3xy+10x-15y\; /\; x^2+2xy+y^2-100}}{}}{}\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\frac{x}{2x+2}\frac{2x}{x^2-2x-3}\frac{x}{6-2x}\frac{5}{-x^2+5x-6}\frac{x+3}{2-x}\frac{3}{x^2+x-2}\frac{1}{x-1}\frac{-4}{x+2}\frac{x}{2x+2}\frac{2x}{x^2-2x-3}\frac{x}{6-2x}\frac{5}{-x^2+5x-6}\frac{x+3}{2-x}\frac{3}{x^2+x-2}\frac{1}{x-1}\frac{-4}{x+2}5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x\; +3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}{\left(x+\frac{1}{5}\right)}^2+\frac{17}{25}=\frac{26}{25}\frac{}{}\left(\frac{7}{x-2}+\frac{8}{x+2}\frac{14x}{x^2-4}\right):\left(\frac{1}{x-2}-\frac{1}{x+2}\right)\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\frac{x-1}{x+2}+\frac{x+3}{x-4}=\frac{2}{\left(x-2\right)\left(4-x\right)}\frac{7x-3}{x-x^3}=\frac{1}{x-1}-\frac{5}{\left(x-1\right)}\frac{1}{x+2}+\frac{1}{x+3}=\frac{1}{\left(x+2\right)\left(x+3\right)}\frac{5}{6x^{2}y}+\frac{7}{12x{y}^2}+\frac{11}{18xy}\frac{4x+2}{15x^{3}y}+\frac{5y-3}{9x^{2}y}+\frac{x+1}{5x{y}^3}\frac{3}{2x}+\frac{3x-3}{2x-1}+\frac{2x^2+1}{4x^2-2x}\frac{x^3+2x}{x^3+1}+\frac{2x}{x^2-x+1}+\frac{1}{x+1}$ \frac{5}{6x^{2}y}+\frac{7}{12x{y}^2}+\frac{11}{18xy}$$ =\frac{30y}{36x^2{y}^2}+\frac{21x}{36x^2{y}^2}+\frac{22xy}{36x^2{y}^2}=\frac{30y+21x+22xy}{36x^2{y}^2}$$ \frac{4x+2}{15x^{3}y}+\frac{}{}$\frac{11x+10}{3x-3}+\frac{15x+13}{4-4x}\frac{5x+3}{x^2-3x}+\frac{9-x}{9-3x}\frac{4xy-1}{5x^{2}y}-\frac{2xy-1}{5x^{2}y}\frac{x+8}{x^2-16}-\frac{2}{x^2+4x}\frac{x^2-49}{2x+1}.\frac{3}{7-x}\frac{3x^2-2x}{x^2-1}.\frac{1-x^4}{{\left(2-3x\right)}^3}\frac{5xy}{2x-3}:\frac{15x{y}^3}{12-8x}\frac{x^2+2x}{3x^2-6x+3}:\frac{2x+4}{5x-5}$