作业帮x+y=0.2,x+3y=1,求3x²+12xy+12y²的值
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/18 20:16:19
xQ;N0< 8(D9( (BD mq̖{8
K4͌Y&,֘EϷkQ~)>nU#.ȟ=.:Q`EYaq(%fC6BZ-Y7 4}eV H4>=/ H IX. Wxq>S>fB z۾4fgQ7&ٸӄfәaRæH>Z=_Pb,d"bM=E K
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