请问求r,59×(1+r)-1+59×(1+r)-2+59×(1+r)-3+59×(1+r)-4+(59+1250)×(1+r)-5=100059×(1+r)-1的-1是次方啊,59×(1+r)-2的-2也是次方,59×(1+r)-3的-3也是次方,59×(1+r)-4的-4都是次方,(1+r)-5的-5
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![请问求r,59×(1+r)-1+59×(1+r)-2+59×(1+r)-3+59×(1+r)-4+(59+1250)×(1+r)-5=100059×(1+r)-1的-1是次方啊,59×(1+r)-2的-2也是次方,59×(1+r)-3的-3也是次方,59×(1+r)-4的-4都是次方,(1+r)-5的-5](/uploads/image/z/2737895-23-5.jpg?t=%E8%AF%B7%E9%97%AE%E6%B1%82r%2C59%C3%97%EF%BC%881%2Br%EF%BC%89-1%2B59%C3%97%EF%BC%881%2Br%EF%BC%89-2%2B59%C3%97%EF%BC%881%2Br%EF%BC%89-3%2B59%C3%97%EF%BC%881%2Br%EF%BC%89-4%2B%EF%BC%8859%2B1250%EF%BC%89%C3%97%EF%BC%881%2Br%EF%BC%89-5%3D100059%C3%97%281%2Br%29-1%E7%9A%84-1%E6%98%AF%E6%AC%A1%E6%96%B9%E5%95%8A%2C59%C3%97%EF%BC%881%2Br%EF%BC%89-2%E7%9A%84-2%E4%B9%9F%E6%98%AF%E6%AC%A1%E6%96%B9%2C59%C3%97%EF%BC%881%2Br%EF%BC%89-3%E7%9A%84-3%E4%B9%9F%E6%98%AF%E6%AC%A1%E6%96%B9%2C59%C3%97%EF%BC%881%2Br%EF%BC%89-4%E7%9A%84-4%E9%83%BD%E6%98%AF%E6%AC%A1%E6%96%B9%2C%EF%BC%881%2Br%EF%BC%89-5%E7%9A%84-5)
请问求r,59×(1+r)-1+59×(1+r)-2+59×(1+r)-3+59×(1+r)-4+(59+1250)×(1+r)-5=100059×(1+r)-1的-1是次方啊,59×(1+r)-2的-2也是次方,59×(1+r)-3的-3也是次方,59×(1+r)-4的-4都是次方,(1+r)-5的-5
请问求r,59×(1+r)-1+59×(1+r)-2+59×(1+r)-3+59×(1+r)-4+(59+1250)×(1+r)-5=1000
59×(1+r)-1的-1是次方啊,59×(1+r)-2的-2也是次方,59×(1+r)-3的-3也是次方,59×(1+r)-4的-4都是次方,(1+r)-5的-5也是次方
请问求r,59×(1+r)-1+59×(1+r)-2+59×(1+r)-3+59×(1+r)-4+(59+1250)×(1+r)-5=100059×(1+r)-1的-1是次方啊,59×(1+r)-2的-2也是次方,59×(1+r)-3的-3也是次方,59×(1+r)-4的-4都是次方,(1+r)-5的-5
此题实际上就是解一个关于r的一元五次方程.
由于一元五次解方程一般没有公式解,而且本题也很难因式分解,
所以只能近似求解.
令f(r)=59(1+r)^-1+59(1+r)^-2+59(1+r)^-3+59(1+r)^-4+(59+1250)(1+r)^-5-1000,则问题就是求f(r)的零点(即令f(r)=0的r)
根据f(r)的图像或者用其导函数,可得出
此f(r)只有一个零点,此零点直接求解较困难,可以根据精确度要求用近似方法(如对分法,迭代法等)求出,
如精度要求|f(r)|