先化简,再求值:9x^3-xy^2/(3x+y)(9x^2-6xy+y^2),其中y/x等于1/3咋做?

问题描述:

先化简,再求值:9x^3-xy^2/(3x+y)(9x^2-6xy+y^2),其中y/x等于1/3咋做?

y/x=1/3
所以x=3y
原式=x(3x+y)(3x-y)/(3x+y)(3x-y)²
=x/(3x-y)
=3y/(3×3y-y)
=3y/8y
=3/8