Решите пожалуйста,очень нужно

$x^2-11x=0\\x(x-11)=0\\x=0\\x-11=0\\x_1=0\\x_2=11$

$16x^2-8x+1=0\\(4x-1)^2=0\\4x-1=0\\4x=1\\x=\frac{1}{4}=0.25$

$x^2-7x+12=0\\x^2-3x-4x+12=0\\x(x-3)-4(x-3)=0\\(x-3)(x-4)=0\\x-3=0\\x-4=0\\x_1=3\\x_2=4$

$2x^4-x^2-3=0\\x^2=t\\2t^2-t-3=0\\t=-1\\t=\frac{3}{2}\\x=-1\\x\notin R\\x=\frac{3}{2}\\x_1=-\frac{\sqrt{6}}{2}\\x_2= \frac{\sqrt{6}}{2}$

$\frac{1}{x}+\frac{2}{x-2}=1,~~x\neq 0,~~x\neq 2\\\frac{1}{x}+\frac{2}{x-2}-1=0\\\frac{x-2+2x-x(x-2)}{x(x-2)}=0\\\frac{x-2+2x-x^2+2x}{x(x-2)}=0\\\frac{5x-2-x^2}{x(x-2)}=0\\5x-2-x^2=0\\-x^2+5x-2=0\\x^2-5x+2=0\\x_{1,2} = \frac{5+-\sqrt{17}}{2}$