先化解,再求值：【1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)】/1/(x2+9x+20),其中x=√2
问题描述：
先化解,再求值：【1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)】/1/(x2+9x+20),其中x=√2
答
原式=[1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)] ×(x+4)(x+5)
=[1/(x+1)-1/(x+2) +1/(x+2)-1/(x+3) +1/(x+3)-1/(x+4)]×(x+4)(x+5)
= [1/(x+1)-1/(x+4)]×(x+4)(x+5)
=3/(x+1)(x+4)×(x+4)(x+5)
=3(x+5)/(x+1)
当x=√2时
=3(√2+5)/(√2+1)
=3(√2+5)(√2-1)
=3(2-√2+5√2-5)
=12√2 -9