Question

当x=根号3/3,求代数式(x^2-3x)/(x-2)/[x+3/(2-x)]的值

Answer 1

原代数式=[(x²-3x)/(x-2)]/[(x²-2x-3)/(x-2)]=[(x²-3x)/(x-2)]/[(x-3)(x+1)/(x-2)]=x/(x+1)

将x=√3/3代入

原式=(√3/3)/(√3/3+1)=(√3-1)/2