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求助消参x=(√3/4)(t1-t2),y=(1/4)(t1+t2),t1^2+t1t2+t2^2=12
人气:380 ℃ 时间:2019-11-15 02:15:38
解答
t1-t2=✓3/4x,t1^2-2t1t2+t2^2=3/16x^2t1+t2=1/4y,t1^2+2t1t2+t2^2=1/16y^2,12+t1t2=1/16y^24t1t2=1/16y^2-3/16x^2t1t2=1/64y^2-3/64x^212+1/64y^2-3/64x^2=1/16y^212-3/64x^2=3/64y^2(1/16x)^...不是啊,我的式子不是x=(√3/4)(t1-t2)吗???你的第一二步为什么是这样?t1^2-2t1t2+t2^2=16x^2/3t1^2+2t1t2+t2^2=16y^2,t1t2+12=16y^24t1t2=16y^2-16x^2/3t1t2=4y^2-4x^2/3=16y^2-1212y^2+4x^2/3=12y^2+x^2/3^2=1
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