已知x=2007,y=2008,求x^2+2xy+y^2/5x^2-4xy除以x+y/5x-4y再+x^2-y/x的值
问题描述:
已知x=2007,y=2008,求x^2+2xy+y^2/5x^2-4xy除以x+y/5x-4y再+x^2-y/x的值
答
原式=(x+y)²/x(5x-4y) x (5x-4y)/(x+y) + (x²-y)/x
=(x+y)/x+ (x²-y)/x
=(x+x²)/x
=1+x
把x=2007,y=2008代入得:原式=1+2007=2008(x+x²)/x不是等于x吗?怎么会等于x+1呢?(x+x²)/x=x(1+x)/x=1+x