【已知:x+y=0.2,x+3y=1,则3x2+12xy+12y2的值为______.】

问题描述:

  已知:x+y=0.2,x+3y=1,则3x2+12xy+12y2的值为______.

  ∵x+y=0.2①,x+3y=1②,   ∴①+②得:2x+4y=1.2,   ∴x+2y=0.6,   ∴3x2+12xy+12y2   =3(x2+4xy+4y2)   =3(x+2y)2   =3×0.62   =1.08.   故答案为:1.08.