Given:
a)
[tex]\begin{gathered} \frac{2x+1}{x}=\frac{1}{x-2} \\ (2x+1)(x-2)=x\times1 \\ 2x^2-4x+x-2=x \\ 2x^2-4x-2=0 \\ \text{Use quadratic formula: ax}^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4\cdot\:2\left(-2\right)}}{2\cdot\:2} \\ x=\frac{-\left(-4\right)\pm\:4\sqrt{2}}{2\cdot\:2} \\ x=1+\sqrt{2},\: x=1-\sqrt{2} \end{gathered}[/tex]
b)
[tex]\begin{gathered} \frac{1}{x+2}=\frac{2}{x-1} \\ x-1=2(x+2) \\ x-1=2x+4 \\ x=-5 \end{gathered}[/tex]
c)
[tex]\begin{gathered} \frac{x+3}{1-x}=\frac{x+1}{x+2} \\ (x+3)(x+2)=(x+1)(1-x) \\ x^2+2x+3x+6=1-x^2 \\ 2x^2+5x+5=0 \\ \text{Use quadratic formula,} \\ x=\frac{-5\pm\sqrt{5^2-4\cdot\:2\cdot\:5}}{2\cdot\:2} \\ x=\frac{-5\pm\sqrt{15}i}{2\cdot\:2} \\ x=\frac{-5+\sqrt{15}i}{2\cdot\:2},\: x=\frac{-5-\sqrt{15}i}{2\cdot\:2} \\ x=-\frac{5}{4}+i\frac{\sqrt{15}}{4},\: x=-\frac{5}{4}-i\frac{\sqrt{15}}{4} \end{gathered}[/tex]
d)
[tex]\begin{gathered} \frac{x+2}{x+8}=\frac{1}{x+2} \\ (x+2)(x+2)=(x+8) \\ x^2+4x+4=x+8 \\ x^2+3x-4=0 \\ x^2-x+4x-4=0 \\ x(x-1)+4(x-1)=0 \\ (x-1)(x+4)=0 \\ x-1=0,x+4=0 \\ x=1,-4 \end{gathered}[/tex]
