已知a^2+4a+1=0,(a^4+ma+1)/(3a^3+ma^2+3a)=5, 則 m= ?
已知a^2+4a+1=0,(a^4+ma+1)/(3a^3+ma^2+3a)=5, 則 m= ?
數(shù)學(xué)人氣：728 ℃時(shí)間：2020-06-12 18:30:42
優(yōu)質(zhì)解答
幾天了,沒(méi)人回答,太不給面子了,用Matlab給個(gè)答案算了.>> [a,m]=solve('a^2+4*a+1=0','(a^4+m*a+1)/(3*a^3+m*a^2+3*a)=5','a,m') a =[ -2+3^(1/2)][ -2-3^(1/2)]m =[ 259/23-37/23*3^(1/2)][ 259/23+37/23*3^(1/2)]...
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