Question

数学题目关于先化简，再求值的

1、（1-x分之2x -x）乘（1+x）²分之1-x²，x=根号2
2、（x²+4xy+4y²分之x²y-4y三次方）乘（x-2y分之4xy +x） ，x=根号2-1 ，y=根号2+1
3、a²-1分之a-2 除以（a-1 -a+1分之2a-1)其中a是方程x²-x=6的根

Answer 1

解1题：
原式=[2x/(1-x) -x]× (1-x²)/(1+x)²
=[2x/(1-x) - x(1-x)/(1-x)]× (1-x)/(1+x)
=[(2x-x+x²)/(1-x)] × (1-x)/(1+x)
=(x+x²)/(1+x)
=x(1+x)/(1+x)
=x
当x=√2时，原式=√2

解2题
原式=[(x²y-4y³)/(x²+4xy+4y²)]×[ 4xy/(x-2y) +x]
=[y(x+2y)(x-2y)/(x+2y)²] ×[ 4xy/(x-2y) +x(x-2y)/(x-2y) ]
=[ y(x-2y)/(x+2y) ]× [(4xy+x²-2xy)/(x-2y) ]
=[ y/(x+2y) ]×(x²+2xy)
=[ y/(x+2y) ]× x(x+2y)
=xy
当x=√2 -1 , y=√2 +1时
原式=(√2 -1)(√2 +1)
=√2²-1²
=2-1
=1

解3题
原式=[(a-2)/(a²-1)] ÷[a-1 -(2a-1)/(a+1)]
=[(a-2)/(a²-1)] ÷[(a-1)(a+1)/(a+1) -(2a-1)/(a+1)]
=[(a-2)/(a²-1)] ÷[(a²-1-2a+1)/(a+1)]
=[(a-2)/(a²-1)] ÷[(a²-2a)/(a+1)]
={ (a-2)/[(a+1)(a-1)] }×{(a+1)/[a(a-2)] }
=1/[a(a-1)]
=1/(a²-a)
因为a是方程x²-x=6的根，把x=a代入方程，得 a²-a=6
当a²-a=6时，原式=1/6

Answer 2