设a>b>c>d>0,且x=√ab+√cd,y=√ac+√bd,z=√ad+√bc,设a>b>c>d>0,且x=√ab+√cd,y=√ac+√bd,z=√ad+√bc,则x,y,z的大小关系为( )A.x
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设a>b>c>d>0,且x=√ab+√cd,y=√ac+√bd,z=√ad+√bc,设a>b>c>d>0,且x=√ab+√cd,y=√ac+√bd,z=√ad+√bc,则x,y,z的大小关系为( )A.x
设a>b>c>d>0,且x=√ab+√cd,y=√ac+√bd,z=√ad+√bc,
设a>b>c>d>0,且x=√ab+√cd,y=√ac+√bd,z=√ad+√bc,则x,y,z的大小关系为( )
A.x
设a>b>c>d>0,且x=√ab+√cd,y=√ac+√bd,z=√ad+√bc,设a>b>c>d>0,且x=√ab+√cd,y=√ac+√bd,z=√ad+√bc,则x,y,z的大小关系为( )A.x
x-y=√ab+√cd-√ac-√bd
=(√ab-√ac)-(√bd-√cd)
=√a(√b-√c)-√d(√b-√c)
=(√a-√d)(√b-√c)>0
=>x>y
排除A,C
y-z=√ac+√bd-√ad-√bc
=(√ac-√ad)-(√bc-√bd)
=√a(√c-√d)-√b(√c-√d)
=(√a-√b)(√c-√d)>0
=>y>z
选D
x2=ab+cd+2√abcd y2=ac+bd+2√abcd z2=ad+bc+2√abcd
x2-y2=a(b-c)+d(c-b)=(a-d)(b-c)
而a>b>c>d>0
所以x2-y2>0 x>y
y2-z2=a(c-d)+b(d-c)= (a-b)(c-d)
而a>b>c>d>0
所以y2-z2>0 y>z
x>y>z选D
x^2-y^2=(a-d)(b-c)>0
x^2-z^2=(a-c)(b-d)>0
y^2-z^2=(a-b)(c-d)>0
所以x>y>z
选D
先 x-y = √ab+√cd-√ac-√bd=√ab-√ac+√cd-√bd
因为a>b>c>d>0,显然√ab>√ac , √cd>√bd
所以 √ab-√ac>0 ,√cd>√bd>0
所以x-y>0 所以 x>y
同理, y-z=√ac+√bd-√ad-√bc>0
所以y>z
所以z