实际问题与一元二次方程

2024-11-22 18:51:30
推荐回答(3个)
回答1:

解:设平均每轮一个人传染x人

1个人传染x人,加上他自己就有( x+1)人,是“乘”不是“加”,第一轮中1个人传染x人,第二轮中还是一个人传染x人,只是传染源变多了。它是翻倍的,懂吗?

请看下例:假如1个人传染5个人
第一轮:1+1×5=6
第二轮:6+6×5=36
(1个人传染5个,加上自己就有6个。 6个人再传染,平均1个传染5个,6个人就是6×5=30,再加上本身的6人。 如果是加6+5(x)=11,你觉得对吗?)
如果你还不懂,我就没办法了,毕竟数学中有些东西是要靠自己理解的,说是说不明白的,好好想想

回答2:

先把方程解出来:设:平均传染为X人。解:1(1+X)平方=121
1(1+X)(1+X)=121 1+2X+X平方=121
X平方+2X-120=0 用一元二次方程解十字相乘法解:
(X+12)(X-10)=0 X1=-12不合题意舍去。 X2=10(人)
答案是传染10人。
你应该理解是:1人得了传染病,传染了10人,1+10=11人;
11人有传染,每人传染10人,那么就是11×10=110人,
11人加上传染的110人,是不是=121人。 这样讲能理解吧。

回答3:

设:……1+(1+x)平方=121解:x=1O

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