已知5x2-3x-5=0,求5x2-2x-(1/5x2-2x-5)的值

已知5x2-3x-5=0,求5x2-2x-(1/5x2-2x-5)的值
已知5x2-3x-5=0,求5x2-2x-(1/5x2-2x-5)的值

已知5x2-3x-5=0,求5x2-2x-(1/5x2-2x-5)的值
This problem can be solved as
To find the value we shall change 5x^2-2x-(1/5x^2-2x-5)
to be 5x^2-3x-5+x+5-(1/5x^2-3x-5+x)
as we know 5x^2-3x-5=0
so the equation can be:x+5-1/x
and x+5-1/x=(x^2+5x-1)/x
takes x^2-1 exchanged by 3x/5
we get (3x/5+5x)/x
cancels x we get
(3x/5+5x)/x= 3/5+5=28/5
Hope it can help you

5x^2=3x+5
原式=x+5-(1/x)
=(x^2+5x-1)/x
=(3x/5+5x-1)/x
=28/5

5x2-3x-5=0
两边+x
5x2-2x-5=x
则5x2-2x=x+5
5x2-3x-5=0
5x2-5=3x
所以x2-1=3x/5
所以原式=x+5-1/x
=(x2+5x-1)/x
=(3x/5+5x)/x
=3/5+5
=28/5

由 5x2-3x-5=0
得 5x2-2x-5=x
5x2-2x=x+5
x2-1=3x/5
∴ 5x2-2x-(1/5x2-2x-5)
=(5x2-2x)-1/(5x2-2x-5)
=x+5-1/x
=(x2+5x-1)/x
=((x2-1)+5x}/x
=(3x/5+5x)/x
=3/5+5
=28/5