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由於1=x2+y2+z2=(x2+12y2)+(12y2+z2)≥2x•y2+2•y2•z=2(xy+yz),當且僅當x=y2=z時,等號成立,∴x=y2=z=12時,xy+yz的最大值為22.故答案為:22.
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