Need help on JJ 173

Hi all,

X^2-Y^2=(X+Y)(X-Y)平方差公式

不是牛人,但好不容易遇到个我会的问题.....顺便做点贡献吧

Another question

180.          1<x<10, 问x的值是几？  B

                （1）x小于5

                （2）(1+ x/100) 的平方的百分位等于它的十分位

比如X=2，则(1+ x/100) 的平方=1.02^2=1.0404. 那么，在1<x<10中，只有当x=6时，百分位等于十分位。故选B

Okay I did what the JJ poster said... did x=6 we have

(1+0.06)^2= 1.06^2= 1.123 I assume 百分位 = 2 and 十分位 = 1 how does that fit the condition of 2?

thanks so this problem goes x^2-Y^2=(x+y)(x-y) how does that help us? what we have here is X^2+y^2

x+y=a, x^ 2+y^2=b, 问xy=？，

so we have (x+y)(x-y)= X^2-Y^2 and since (x+y)= a we can do a(x-y)=X^2-Y^2 then what? Sorry maybe I am not seeing things... brain kida slow after 9 pm

如果要解决这个问题的话用的不是 "平方差公式"啦......||||||汗一个~~

(X+Y)^2=X^2+2XY+Y^2=a^2   由题 X^2+Y^2=b 可得 2XY= a^2-b 然后XY就得出来啦~~