初二一道二次根式题!

2个回答

  • 原式=根号{a²+a²-b²+2a*根号(a²-b²)+[根号(a²-b²)]²-[根号(a²-b²)]²}-根号{a²+b²-b²-2b*

    根号(a²-b²)+[根号(a²-b²)]²-[根号(a²-b²)]²}

    =根号{[a+根号(a²-b²)]²}-根号{[b-根号(a²-b²)]²}

    =|a+根号(a²-b²)|-|b-根号(a²-b²)|

    ∵a>(根号2)*b>0

    ∴a²>2b²>0

    即a²-b²>b²>0

    a+根号(a²-b²)>0 b-根号(a²-b²)<0

    ∴|a+根号(a²-b²)|=a+根号(a²-b²)

    |b-根号(a²-b²)|=[根号(a²-b²)]-b

    ∴原式=a+根号(a²-b²)-{[根号(a²-b²)]-b}

    =a+根号(a²-b²)-[根号(a²-b²)]+b

    =a+