Question

已知4x²+y²-4x-6y+10=0,求[(2/3)x√(9x)+y²√(x/y³)-(x²√(1/x)-5x√(y/x)的值

Answer 1

即(4x²-4x+1)+(y²-6y+9)=0
(2x-1)²+(y-3)²=0
所以2x-1=y-3=0
x=1/2,y=3

原式=2x√x+√(xy)-x√x-5√(xy)
=x√x-4√(xy)
=√2/4-4√(3/2)
=√2/4-2√6

Answer 2

4x²+y²-4x-6y+10=0
4x²-4x+1 +y²-6y+9=0
(2x-1)²+(y-3)²=0
平方项恒非负，两平方项之和=0，两平方项分别=0
2x-1=0 x=1/2
y-3=0 y=3
(2/3)x√(9x)+y²√x/y³-(x²√(1/x) -5x√(y/x)
=2x√x+√(xy)-x√x -5√(xy)
=x√x -4√(xy)
=(1/2)√(1/2) -4√[(1/2)×3]
=√2/4 -2√6

Answer 3

4x²+y²-4x-6y+10=0
4x²-4x+1+y²-6y+9=0
(2x-1)²+(y-3)²=0
(2x-1)²=0,(y-3)²=0
x=1/2,
y=3

[(2/3)x√(9x)+y²√(x/y³)-(x²√(1/x)-5x√(y/x)
=[(2/3)x√(9x)+y√(x/y)-(x√(x²/x)-5√(x²y/x)
=[(2/3)x√(9x)+√(xy²/y)-(x√x)-5√(xy)
=2x√x+√(xy)-(x√x)-5√(xy)
=x√x-4√(xy)
=1/2*√(1/2)-4√(1/2*3)
=√2/4-4√(3/2)
=√2/4-2√6