4q^3-3q-1=04q^3-(4q-q)-1=04q^3-4q+(q-1)=04q(q^2-1)+(q-1)=04q(q+1)(q-1)+(q-1)=0(q-1)[4q(q+1)+1]=0q-1=0,==>q=14q(q+1)+1=0,==>4q^2+4q+1=0,==>(2q+1)^2=0,==>q=-1/2
解方程4q^3-3q-1=0希望能有具体过程,实在不行只有答案也可以
4q^3-3q-1=04q^3-(4q-q)-1=04q^3-4q+(q-1)=04q(q^2-1)+(q-1)=04q(q+1)(q-1)+(q-1)=0(q-1)[4q(q+1)+1]=0q-1=0,==>q=14q(q+1)+1=0,==>4q^2+4q+1=0,==>(2q+1)^2=0,==>q=-1/2