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標題:

maths

發問:

solve 7t^2 - t^3 -36=0 with procedure

最佳解答:

7t2-t3-36=0 t3-7t2+36=0 t3+23-7t2+28=0 (←36=8+28) (t+2)(t2-2t+4)-7(t2-22)=0 (t+2)(t2-2t+4)-7(t+2)(t-2)=0 (t+2)[t2-2t+4-7(t-2)]=0 (t+2)(t2-9t+18)=0 (t-6)(t-3)(t+2)=0 .·.t=-2 or 3 or 6

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7t^2 - t^3 -36=0 t^3-7t^2+36=0 (t-6)(t^2-t-6)=0 (t-6)(t-3)(t+2)=0 t=-2,3or6
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