x+y=0.2,x+3y=1,求3x²+12xy+12y²的值

x+y=0.2,x+3y=1,求3x²+12xy+12y²的值
x+y=0.2,x+3y=1,求3x²+12xy+12y²的值

x+y=0.2,x+3y=1,求3x²+12xy+12y²的值
3x²+12xy+12y²
=3(x²+4XY+4y²)
=3(X+2Y)²
x+y=0.2,(1)x+3y=1(2)
(2)式-（1）,得
2Y=0.8
Y=0.4
X=-0.2
原式=3（-0.2+0.8）²
=1.08

x+y=0.2,
x+3y=1
两式相减得
2y=0.8
y=0.4
x+0.4=0.2
x=-0.2
3x²+12xy+12y²
=3(x²+4xy+4y²)
=3(x+2y)²
=3*(-0.2+0.4*2)²
=3*0.6²
=3*0.36
=1.08

0.16

3x²+12xy+12y²
=3（x²+4xy+4y²）
=3（x+2y）²
=3[1/2（2x+4y)]²
=3/4[（x+y）+（x+3y）]²
=3/4(0.2+1)²
=1.08