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令x1 = y1+y2, x2 = y1-y2, x3 = y3+y4, x4 = y3-y4
得 (1/2)f = (y1)^2 - (y2)^2 - 3(y1+y2)(y3+y4) - 3(y1-y2)(y3-y4) + (y3)^2 - (y4)^2
= (y1)^2 - (y2)^2 - 6y1y3 - 6y2y4 + (y3)^2 - (y4)^2
= (y1-3y3)^2 - (y2+3y4)^2 - 10(y3)^2 + 8(y4)^2
= (z1)^2 - (z2)^2 -10(z3)^2 + 8(z4)^2
f = 2(z1)^2 - 2(z2)^2 -20(z3)^2 + 16(z4)^2
得 (1/2)f = (y1)^2 - (y2)^2 - 3(y1+y2)(y3+y4) - 3(y1-y2)(y3-y4) + (y3)^2 - (y4)^2
= (y1)^2 - (y2)^2 - 6y1y3 - 6y2y4 + (y3)^2 - (y4)^2
= (y1-3y3)^2 - (y2+3y4)^2 - 10(y3)^2 + 8(y4)^2
= (z1)^2 - (z2)^2 -10(z3)^2 + 8(z4)^2
f = 2(z1)^2 - 2(z2)^2 -20(z3)^2 + 16(z4)^2
更多追问追答
追问
为什么要令x3=y3+y4呢,具体是怎么想的
追答
令 x3 = y3+y4, x4 = y3-y4, 会产生(y3)^2, (y4)^2 项,以便配方。
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