x>y,xy=1,求证(x.x+y.y)/(x-y)≥根号8
4个回答
x2+y2=(x-y)2-2xy=(x-y)2-2
原式=(x-y)+2/(x-y)≥2√2
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化简:[(x×根号x+x×根号y)/(xy-y²)]-[(x+根号下xy+y)/(x×根号x-y×根号y)]
1\x+1\y=8\x+y,则y\xy+x\xy=?
化简求值:(X×根号X+X×根号y)÷(XY-Y²)-(X+根号XY+Y)÷X×根号X-Y×根号Y,其中X=1
{化简} [x+y/(根号x+根号y)]+[2xy/x·根号y+y·根号x]