(1-x/3-x)平方除以x平方-6x+9/9-x平方*1/x平方-2x+1

网友回答

格式有点乱,
〖((1-x)/(3-x))^2/((x^2-6x+9)/(9-x^2 ))〗^ ×1/(x^2-2x+1)
=〖((1-x)/(3-x))^2/(〖(X-3)〗^2/((3-x)(3+x)))〗^ ×1/〖(x-1)〗^2 
=〖(1-x)〗^2/〖(3-x)〗^2 *((3-x)(3+x))/(X-3)^2 *1/〖(x-1)〗^2 
=1/〖(3-x)〗^2 *((3-x)(3+x))/(3-x)^2 *〖(1-x)〗^2/〖(x-1)〗^2 
=1/〖(3-x)〗^2 *((3+x))/(3-x)^ *1
=((3+x))/〖(3-x)〗^3
(1-x/3-x)平方除以x平方-6x+9/9-x平方*1/x平方-2x+1(图1)
======以下答案可供参考======
供参考答案1:
题目没写清楚,连猜带蒙,是不是这样的?
[(1-x)/(3-x)]²÷[(x²-6x+9)/(9-x²)]×[1/(x²-2x+1)]
=[(1-x)/(3-x)]²×[(3+x)(3-x)/(x-3)²]×[1/(x-1)²]
=(3+x)/[(3-x)(x-3)²]=(3+x)/(3-x)³