Question

初二的二次根式，大家看一看~~~

计算：3/a[(√a^2/b^2）÷（3a/b^2)]
[(x√y)-(y√x)/(x√y)+(y√x)]-[(y√x)+(x√y)/(y√x)-(x√y)]
麻烦大家也看看这个，也是计算。

Answer 1

3/a[(√a^2/b^2）÷（3a/b^2)]=3/a[|a|/|b|*b^2/3a]=3*|a|*|b|/3=|a*b|
[(x√y)-(y√x)/(x√y)+(y√x)]-[(y√x)+(x√y)/(y√x)-(x√y)]
={√(xy)(√x-√y)/[√(xy)(√x+√y)]}-{√(xy)(√x+√y)/[√(xy)(√x-√y)]}
=(√x-√y)/(√x+√y)-(√x+√y)/(√x-√y)
=(√x-√y)²/(x-y)-(√x+√y)²/(x-y)
=[(√x-√y)²-(√x+√y)²]/(x-y)
=-4√(xy)/(x-y)