先化简再求值(3x-2x²)(3-2x-x²)/(x²+x)(2x²-5x+3),其中x=【2】怎么写啊,根本化简不鸟!

(3x-2x²)(3-2x-x²)/(x²+x)(2x²-5x+3)=x(3-2x)(3+x)(1-x)/x(x+1)(2x-3)(x-1)
=(x+3)/(x+1)=5/3原来如此啊，我忘了十字相乘法，怪不得算到x(3-2x)(3-2x-x2）/x(x+1)(2x2-5x+3)谢谢