Знатоки алгебры сюда!

10)
$\frac{3x-7}{x+5}=\frac{x-3}{x+2};\\ \frac{3x-7}{x+5}-\frac{x-3}{x+2}=0;\\ \frac{(3x-7)\cdot(x+2)-(x-3)\cdot(x+5)}{(x+3)\cdot(x+5)}=0;\\ D(f): \left \{ {{x\neq-3} \atop {x\neq-5}} \right. \ \ \ x\in(-\infty;-5)\bigcup(-5;-3)\bigcup(-5;+\infty);\\ 3x^2-7x+6x-14-x^2-5x+3x+15=0;\\ 2x^2-3x+1=0;\\ D=9-8=1;\\ x_1=\frac{3-1}{4}=-\frac12;\\ x_2=\frac{3+1}{4}=1;\\$
$1:\frac{3-7}{1+5}=\frac{-4}{6}=-\frac23;\ \ \ \frac{1-3}{1+2}=-\frac{-2}{3}=-\frac23;\\ \frac{3x-7}{x+5}=\frac{x-3}{x+2}|_{x=1};\\ \\ -\frac12:\ \ \ \ \frac{\frac32-7}{\frac12+5}=\frac{\frac{3-14}{2}}{\frac{1+10}{2}}=\frac{3-14}{1+10}=\\ =-\frac{11}{11}=-1;\\ \frac{\frac12-3}{\frac12+2}=\frac{\frac{1-6}{2}}{\frac{1+4}{2}}=\frac{1-6}{1+4}=\frac{-5}{5}=-1$

11)x_2=3." alt="x^2-2x-a=0;\\ x_1x_2=3." align="absmiddle" class="latex-formula">
теорема Виетта
$x^2-x-20=0;\\ x_1+x_2=1;\\ x_1\cdotx_2=20;\\ x_1=-4;\\ x_2=5$