4X^2+Y^2-4X-6Y+10=0
(4X^2-4X+1)+(Y^2-6Y+9)=0
(2X-1)^2+(Y-3)^2=0
平方大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立
所以两个都等于0
所以2X-1=0,Y-3=0
X=1/2,Y=3
[2/3X√(9X)+Y^2√(X/Y^3)]-[X^2√(1/X)-5X√(Y/X)]
=[1/3*√(9/2)+9*√(1/54)]-[(1/4)√2-5/2*√6]
=√2/2+√6/2-√2/4+5/2*√6
=√2/4+3√6
√5=2.236
[√80-√(9/5)]-[√(16/5)+4/5√45]
=4√5-3/5*√5-4/5*√5-12/5*√55
=0.2*√5
=0.2*2.236
=0.4472
(1-2√3)(1+2√3)-(2√3-1)^2
=1^2-(2√3)^2-[(2√3)^2-4√3+1]
=1-12-12+4√3-1
=-24+4√3