solving and finding c1 c2 values

2020 4 16
1 个回答
更新时间：2020 4 16
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Eq4 =
(2682754076831727*c1)/8388608 + (2682754076831727*c2)/8388608 + (2682754076831727*c3)/8388608 + (3167144087592971*(c1 + c2 + c3)^3)/55083008 - 24
Eq5 =
(2682754076831727*c1)/8388608 + (2682754076831727*c2)/8388608 + (2682754076831727*c3)/8388608 + (3167144087592971*(c1 + c2 + c3)^3)/55083008 - 24
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回答（1 个）

Star Strider
Star Strider 2020-4-16

0 个投票

Try this:
syms c1 c2 c3
Eq4 = (2682754076831727*c1)/8388608 + (2682754076831727*c2)/8388608 + (2682754076831727*c3)/8388608 + (3167144087592971*(c1 + c2 + c3)^3)/55083008 - 24;
Eq5 = (2682754076831727*c1)/8388608 + (2682754076831727*c2)/8388608 + (2682754076831727*c3)/8388608 + (3167144087592971*(c1 + c2 + c3)^3)/55083008 - 24;
S = solve(Eq4,Eq5);
c1 = S.c1
c2 = S.c2
Note that ‘c1’ is a function of ‘c3’, so unlless ‘c3’ has a numeric value or an expression of its own that would provide a numeric value for it, a strictly numeric solution is not possible.

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