y=(1+x)^2/[x(x-2)^2],x∈(0,1/2),
y'=[2(1+x)*x(x-2)^2-(1+x)^2*(3x^2-8x+4)]/[x(x-2)^2]^2
=(1+x)(2x^3-8x^2+8x
-3x^3+8x^2-4x
-3x^2+8x-4)/[x(x-2)^2]^2
=(x+1)(-x^3-3x+12x-4)/[x(x-2)^2]^2
=-(x+1)(x^2+5x-2)/[x^2*(x-2)^3]
=-(x+1)[x-(-5-√33)/2][x-(-5+√33)/2]/[x^2*(x-2)^3],
當(dāng)x=(-5+√33)/2時y取最小值=(3-5x)/[x(6-9x)](利用x^2=2-5x)
=(3-5x)/(51x-18)
=(31-5√33)/(-291+51√33)
=(31-5√33)(291+51√33)/1152
=(606+126√33)/1152
=(101+21√33)/192.抱歉, 題目弄錯了, 能幫忙再做一下么,?
y=x(1-2x)分之(1+x)平方,x∈(0,二分之一),求最小值y=(1+x)^2/[x(1-2x)],
y'=[2(1+x)(x-2x^2)-(1-4x)(1+x)^2]/[x(1-2x)]^2
=(1+x)(2x-4x^2-1+3x+4x^2)/[x(1-2x)]^2
=5(x+1)(x-1/5)/[x(1-2x)]^2,
x=1/5時y取最小值（36/25）/（3/25）=12.