Question

数学 化简求值

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

Answer 1

x+y=0.2,x+3y=1

(x+y)+(x+3y)=0.2+1
2x+4y=1.2
x+2y=0.6

3x²+12xy+12y²

= 3(x^2+4xy+4y^2)
=3(x+2y)^2
=3*0.6^2
=3*0.36
=1.08

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Answer 2

3x²+12xy+12y²
=3(x^2+4xy+4y^2)
=3(x+2y)^2

x+y=0.2,
x+3y=1
2y=0.8
y=0.4
x=-0.2

3(x+2y)^2
=3(-0.2+0.8)^2
=3*0.36
=1.08

Answer 3

x+y=0.2 (1)
x+3y=1 (2)
把(1)式(2)式左右两边分别相加，得2x+4y=1.2 即 x+2y=0.5

3x²+12xy+12y²=3(x²+4xy+(2y)²)=3(x+2y)²

所以，原式=3X0.5²=3X0.25=0.75

Answer 4

=3（x²+4xy+4y²）
=3（x+2y)²
{ x+y=0.2
{x+3y=1
相减2y=0.8
y=0.4
将Y=0.4带入x+y=0.2得
x=-0.2
所以3(x+2y）²=3（-0.2+2×0.4）=1.8

Answer 5

解：原式=3（x²+4xy+3y²)=3(x+y)(x+3y)=3*0.2*1=0.6

Answer 6

最终答案——————76.5