x^2+x-1=0 求x(1-(2/3-x))/(x+1)
问题描述：
x^2+x-1=0 求x(1-(2/3-x))/(x+1)
急
答
x^2+x-1=0,x1=(√5-1)/2,x2=(-√5-1)/2.且 x^2=1-x.
f(x)=x[1-2/(3-x)]/(1+x) = x(1-x)/[(3-x)(1+x)]
= (x-x^2)/(3+2x-x^2) = (2x-1)/(3x+2)
= (2x+4/3-7/3)/(3x+2) = 2/3-(7/3)/(3x+2)
则 f(x1) = 2/3-(7/3)(2/(3√5+1)= 2/3-(7/66)(3√5-1)=(17-7√5)/22.
f(x2) = 2/3-(7/3)(2/(-3√5+1)= 2/3+(7/66)(3√5+1)=(17+7√5)/22.
得所求的值是 (17±7√5)/22.