八年级上数学，三道计算求解

Answer 1

   原式=[根号(x+1)-根号(x-1)]^2/[根号(x+1)+根号(x-1)][根号(x+1)-根号(x-1)]

          =[x+1+x-1-2根号(x^2-1)]/[(x+1-(x-1)]

          =[2x-2根号(x^2-1)]/2

          =x-根号(x^2-1).

  2x^2+3x-7=0

    x^2+(3/2)x=7/2

    x^2+(3/2)x+(9/16)=(7/2)+(9/16)

     (x+3/4)^2=65/16

     x+3/4=正负(根号65)/4

     x1=(-3+根号65)/4,      x2=(-3-根号65)/4.

    原式=[1除以(2/5)X4]根号[(10/3)除以(7/3)X(7/5)]

           =10根号2.

Answer 2

[√(x+1)-√(x-1)]/[√(x+1)+√(x-1)]
=[√(x+1)-√(x-1)][√(x+1)-√(x-1)]/{[√(x+1)+√(x-1)]/[√(x+1)-√(x-1)]}
={[√(x+1)]²-2√(x-1)*√(x+1)+[√(x-1)]²}/(x+1-x+1)
=[x+1-2√(x²-1)+x-1]/2
=x-√(x²-1)

2x²+3x-7=0
x²+3/2x=7/2
x²+3/2x+9/16=7/2+9/16
(x+3/4)²=65/16
x+3/4=±√(65/16)
x=(-3±√65)/4

√3又1/3÷2/5√2又1/3×4√1又2/5
=√10/3÷2/5√7/3×4√7/5
=（4÷2/5)×√(10/3÷7/3×7/5)
=10√2