题目

求当X^2+2X-2=0时,分式(X^2-4/X^2+X+1)^2/(X^3-2X^2/X^3+X^2+X)^2*X/(X+2)^3的值

提问时间：2020-06-25

答案

∵ x²+2x-2=0
∴ x²+2x=2
原式=[(x²-4)/(x²+x+1)]²÷[(x³-2x²)/(x³+x²+x)]²×x/(x+2)³
=[(x+2)(x-2)/(x²+x+1)]²÷{x²(x-2)/[x(x²+x+1)]}²×x/(x+2)³
=[(x+2)²(x-2)²/(x²+x+1)²]÷{x^4(x-2)²/[x²(x²+x+1)²]}×x/(x+2)³ ( ^ 表示乘方 )
=[(x+2)²(x-2)²/(x²+x+1)²]×{x²(x²+x+1)²/[x^4(x-2)²]}×x/(x+2)³
=1/[x(x+2)]
=1/(x²+2x)
=1/2