Question

1/x+2/x+1=2
"1/x";"2/x+1" - дроби

Answer 1

$\frac{1}{x}+ \frac{2}{x+1}=2 \\ \frac{1*(x+1)+2x}{x(x+1)}=2 \\ \frac{3x+1}{x(x+1)}=2 \\ x(x+1) \neq 0,x \neq 0,x \neq -1 \\ 3x+1=2x(x+1) \\ 3x+1=2x^2+2x \\ 2x^2-x-1=0 \\ D=1^2-4*2*(-1)=1+8=9 \\ x_1= \frac{1+ \sqrt{D} }{2*2}= \frac{1+ \sqrt{9}}{4}=1 \\ x_2=\frac{1-\sqrt{D} }{2*2}= \frac{1-\sqrt{9}}{4}=-0.5$