she is a student () No.1 Middle School.A:in ,N:of C:at

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• 设a为任意有理数,b为何值时有理系数方程有有理根设a为任意有理数,b为何值时有理系数方程2x^2+(a+1)x-(3a^2-4a+b)=0的根是有理数Δ=(a+1)^2-4*2*[-(3a^2-4a+b)]=a^2+2a+1+24a^2-32a+8b=25a^2-30a+8b+1=(5a-3)^2+8b-8因为根是有理数,a为任意有理数所以8b-8=0所以b=1或设 25a^2-30a+8b+1=k^2b=(k^2-25a^+30a-1)/8因为k为有理数a为有理数 所以k^2-25a^+30a-1为有理数有理数除以有理数还为有理数,所有b为任意有理数请问那个答案错 为什么不能说因为第一个对就说第二个错，必须详细说为什么 第二个也有道理 a与k也都可以取任意有理数，b也是任意的啊 代数到第二种解法肯定是错的 但否掉这种解法不能只代数 第二种解法还是有道理的 他本身是不成立的 要证明第二种解法没道理
• 在有理数范围内定义&其规则是a&b=a-b^2根据这个规则,方程（2x+1）&3=0的解是______-