设实数a、b,分别满足19a^2+99b+1=0,b^2+99b+19=0.并且ab≠1,(ab+4a+1)÷b的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/28 10:56:55
![设实数a、b,分别满足19a^2+99b+1=0,b^2+99b+19=0.并且ab≠1,(ab+4a+1)÷b的值](/uploads/image/z/12975671-47-1.jpg?t=%E8%AE%BE%E5%AE%9E%E6%95%B0a%E3%80%81b%2C%E5%88%86%E5%88%AB%E6%BB%A1%E8%B6%B319a%5E2%2B99b%2B1%3D0%2Cb%5E2%2B99b%2B19%3D0.%E5%B9%B6%E4%B8%94ab%E2%89%A01%2C%EF%BC%88ab%2B4a%2B1%29%C3%B7b%E7%9A%84%E5%80%BC)
设实数a、b,分别满足19a^2+99b+1=0,b^2+99b+19=0.并且ab≠1,(ab+4a+1)÷b的值
设实数a、b,分别满足19a^2+99b+1=0,b^2+99b+19=0.并且ab≠1,(ab+4a+1)÷b的值
设实数a、b,分别满足19a^2+99b+1=0,b^2+99b+19=0.并且ab≠1,(ab+4a+1)÷b的值
将19+99b+b^2=0,除以b^2,得1+99(1/b)+19(1/b)^2=0.可知,a,1/b是19x^2+99x+1=0的两不等根,(若相等ab=1,矛盾),由韦达定理,a+1/b=-99/19,a/b=1/19.(ab+4a+1)/b=a+1/b+4a/b=-99/19+4*1/19==-5