Đáp án:
$\rm x^2-(2m-1)x+m^2-m=0\\\text{PT có 2 nghiệm}\\↔\Delta \geq 0\\↔4m^2-4m+1-4(m^2-m) \geq 0\\↔1>0(\text{luôn đúng})\\\sqrt{x_1}=\sqrt{2x_2}(x_1,x_2 \geq 0)\\↔x_1=2x_2\\Vì\,\,x_1,x_2 \geq 0\\\to x_1.x_2,x_1+x_2 \geq 0\\\to 2m-1 \geq 0,m^2-m \geq 0\\\to m \geq \dfrac{1}{2},m(m-1) \geq 0\\\to m \geq 1\\x_1=2x_2\\\to x_1+x_2=2m-1\\↔3x_2=2m-1\\↔x_2=\dfrac{2m-1}{3},x_1=\dfrac{2(2m-1)}{3}\\\to x_1.x_2=\dfrac{2(2m-1)^2}{9}=m^2-m\\\to 2(4m^2-4m+1)=9m^2-9m\\↔8m^2-8m+2=9m^2-9m\\↔m^2-m-2=0\\↔m_1=-1(l),m_2=(tm)\\Vậy\,\,m=2\,\,thì\,\,\sqrt{x_1}=\sqrt{2x_2}$
