Question

решите умоляю 13.11 б

Answer 1

  $left { {{y=sqrt{x}} atop {y=2x-1}} right.; left { {{y=sqrt{x}} atop {sqrt{x}=2x-1}} right.\\2x-sqrt{x}-1=0; ,x geq 0 \t=sqrt{x} geq 0\2t^2-t-1=0\D=1+8=9\t_1=1 geq 0, t_2=-frac{1}{2}<0; -; ne; podxodit\sqrt{x}=1to x=1$