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化二次型(x1+x2)^2+(x2-x3)^2+(x3+x1)^2为标准型,
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化二次型(x1+x2)^2+(x2-x3)^2+(x3+x1)^2为标准型,
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答案和解析
f=2x1^2+2x2^2+2x3^2+2x1x2+2x1x3-2x2x3
= 2(x1+(1/2)x2+(1/2)x3)^2+(3/2)x2^2+(3/2)x3^2-3x2x3
= 2(x1+(1/2)x2+(1/2)x3)^2+(3/2)(x2-x3)^2
= 2y1^2+(3/2)y2^2
= 2(x1+(1/2)x2+(1/2)x3)^2+(3/2)x2^2+(3/2)x3^2-3x2x3
= 2(x1+(1/2)x2+(1/2)x3)^2+(3/2)(x2-x3)^2
= 2y1^2+(3/2)y2^2
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