求大神解一下。 5
2个回答
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第二个好解,x=0和3/7;第一个要用三次方求解公式,因为你如果要强行找平公式很麻烦,21的开方很不好找,它的答案是
x1=1/6*(2276+36*3997^(1/2))^(1/3)+2/3/(2276+36*3997^(1/2))^(1/3)+1/3
x2=-1/12(2276+36*3997^(1/2))^(1/3)-1/3/(2276+36*3997^(1/2))^(1/3)+1/3+1/2*i*3^(1/2)*(1/6*(2276+36*3997^(1/2))^(1/3)-2/3/(2276+36*3997^(1/2))^(1/3))
x3=-1/12*(2276+36*3997^(1/2))^(1/3)-1/3/(2276+36*3997^(1/2))^(1/3)+1/3-1/2*i*3^(1/2)*(1/6*(2276+36*3997^(1/2))^(1/3)-2/3/(2276+36*3997^(1/2))^(1/3))
只有x1是实根。你去百度一元三次方求解公式,里面有详细的说明,非常麻烦。
x1=1/6*(2276+36*3997^(1/2))^(1/3)+2/3/(2276+36*3997^(1/2))^(1/3)+1/3
x2=-1/12(2276+36*3997^(1/2))^(1/3)-1/3/(2276+36*3997^(1/2))^(1/3)+1/3+1/2*i*3^(1/2)*(1/6*(2276+36*3997^(1/2))^(1/3)-2/3/(2276+36*3997^(1/2))^(1/3))
x3=-1/12*(2276+36*3997^(1/2))^(1/3)-1/3/(2276+36*3997^(1/2))^(1/3)+1/3-1/2*i*3^(1/2)*(1/6*(2276+36*3997^(1/2))^(1/3)-2/3/(2276+36*3997^(1/2))^(1/3))
只有x1是实根。你去百度一元三次方求解公式,里面有详细的说明,非常麻烦。
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