x²+3x-1=0x²+3x=1(x-3)/(3x^2-6x)除以[x+2-5/(x-2)]=(x-3)/3x(x-2)÷(x²-4-5)/(x-2)=[(x-3)/3x(x-2)]*(x-2)/(x-3)(x+3)=1/3x(x+3)=1/3x²+9x=1/3(x²+3x)=1/3