Question

已知x+y=-8，xy=8，求【y根号x分之y】+【x根号y分之x】的值

Answer 1

方法如下，
请作参考：

追答

X<0，y<0，应为-12√2

Answer 2

解：由X+y=-8,Xy=8知：X<0且y<0
y√(y/X)+X√(X/y)
=y√(Xy/X^2)+X√(Xy/y^2)
=-y/X√Xy-X/y√Xy
=-√Xy(y/X+X/y)
=-√Xy[(X^2+y^2)/Xy]
=-(X^2+y^2+2Xy-2Xy)/√Xy
=-[(X+y)^2-2Xy]/√Xy
=-[(-8)^2-2×8]/√8
=-48/√8
=-6√8

Answer 3

先化简y√(y/x)+x√(x/y)=y(√xy)/x+x(√xy)/y
=(√xy)(y/x+x/y)=(√xy)(x^2+y^2)/xy
=[(x+y)^2-2xy]/(√xy),
代入=【64-16】/√8=48/(2√2)=12√2

Answer 4

∵x+y=-8且xy=8
∴x² + y²=(x+y)²-2xy
=(-8)² - 2×8=48
则原式=y√(xy/x²) + x√(xy/y²)
=(y/x)·√xy + (x/y)·√xy
=(y/x + x/y)·√xy
=[(y²+x²)/xy]·√xy
=(48/8)·√8
=6·2√2
=12√2