√y^3-6xy^2+9x^2y/x(y
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√y^3-6xy^2+9x^2y/x(y
√y^3-6xy^2+9x^2y/x(y<3x<0)
rt 整个式子都是在根号下的
√y^3-6xy^2+9x^2y/x(y
即y-3x<0
x<0
所以(y-3x)/x>0
原式=√[y(y-3x)²/x]
=√[xy(y-3x)²/x²]
=√(xy)*|(y-3x)/x]
=(y-3x)√(xy)/x
y<3x<0
y-3x<0 x<0
于是
√[(y^3-6xy^2+9x^2y)/x]
=√[y(y²-6xy+9x²)/x]
=√[y(y-3x)²/x]
=√[xy(y-3x)²/x²]
=(y-3x)/x*√(xy)
化简:x-y/x+3y÷x^-y^2/x^2+6xy+9y^2-(xy/x+y)
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