Question

решите уравнения, 30 баллов!

Answer 1

$\frac{5}{y+3}-\frac{3}{y}=\frac{2-y}{y^2+3y} \: \: \: \: \: | \:y \ne -3, y\ne0\\ \frac{5}{y+3}-\frac{3}{y}-\frac{2-y}{y^2+3y}=0\\\frac{5}{y+3}-\frac{3}{y}-\frac{2-y}{y(y+3)}=0\\\frac{5y-3(y+3)-(2-y)}{y(y+3)}=0\\\frac{5y-3y-9-2+y}{y(y+3)}=0\\\frac{3y-11}{y(y+3)}=0\\3y-11=0\\3y=11\\y=\frac{11}{3}\\y=3\frac{2}{3}$

$\frac{5}{x-2}+1=\frac{14}{x^2-4x+4} \: \: \: \: \: | \: x\ne2\\\frac{5}{x-2}+1-\frac{14}{(x-2)^2}=0\\\frac{5(x-2)+(x-2)^2-14}{(x-2)^2}=0\\\frac{5x-10+(x-2)^2-14}{(x-2)^2}=0\\\frac{5x-24+x^2-4x+4}{(x-2)^2}=0\\\frac{x-20+x^2}{(x-2)^2}=0\\x-20+x^2=0\\x^2+x-20=0\\D=1^2-4\cdot1\cdot(-20)=1+80=81\\\sqrt{D}=\sqrt{81}=9\\x_1=\frac{-1+9}{2}=\frac{8}{2}=4\\x_2=\frac{-1-9}{2}=\frac{-10}{2}=-5$