Question

$\frac{40}{x-18} - \frac{40}{x+2}=1$

Answer 1

$\frac{40}{x-18} -\frac{40}{x+2}=1\\\\\frac{40x+80-40x+720}{(x-18)(x+2)}=1\\\\\frac{800}{(x-18)(x+2)}=1\\\\(x-18)(x+2)=800\\\\x\neq18;x\neq-2\\\\x^{2}+2x-18x-36-800=0\\\\x^{2} -16x-836=0\\\\\frac{D}{4}=8^{2}+836=64+836=900=30^{2} \\\\x_{1}=8+30=38\\\\x_{2}=8-30=-22$

Answer 2

$\frac{40}{x-18}-\frac{40}{x+2}=1\; \; ,\; \; \; ODZ:\; x\ne 18\; ,\; x\ne -2\\\\\frac{40(x+2)-40(x-18)-(x-18)(x+2)}{(x-18)(x+2)}=0\\\\\frac{40x+80-40x+720-(x^2-16x-36)}{(x-18)(x+2)}=0\\\\\frac{-x^2+16x+836}{(x-18)(x+2)}=0\\\\x^2-16x-836=0\; \; ,\; \; D/4=8^2+836=900\; ,\\\\x_1=8-30=-22\; \; ,\; \; x_2=8+30=38\\\\Otvet:\; \; -22\; ,\; \; 38\; .$