Question

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

Answer 1

$frac{40}{x-18} -frac{40}{x+2}=1\\frac{40x+80-40x+720}{(x-18)(x+2)}=1\\frac{800}{(x-18)(x+2)}=1\\(x-18)(x+2)=800\\xneq18;xneq-2\\x^{2}+2x-18x-36-800=0\\x^{2} -16x-836=0\\frac{D}{4}=8^{2}+836=64+836=900=30^{2} \\x_{1}=8+30=38\\x_{2}=8-30=-22$

Answer 2

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