已知x+y=0.2,x+3y=1,求3x²+12xy+12y²的值
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已知x+y=0.2,x+3y=1,求3x²+12xy+12y²的值
![已知x+y=0.2,x+3y=1,求3x²+12xy+12y²的值](/uploads/image/z/7571887-7-7.jpg?t=%E5%B7%B2%E7%9F%A5x%2By%3D0.2%2Cx%2B3y%3D1%2C%E6%B1%823x%26%23178%3B%2B12xy%2B12y%26%23178%3B%E7%9A%84%E5%80%BC)
x+y=0.2,x+3y=1,2y=0.8 x=-0.2
3x²+12xy+12y²=3(x²+4xy+4y²)=3(x+2y)²=3*0.6²=1.08
3x²+12xy+12y²=3(x²+4xy+4y²)=3(x+2y)²=3*0.6²=1.08
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