y = (x-5)(x^2-2x+1) / (x-7)(x^2+2x+3)
= (x-5)(x-1)(x-1) / (x-7)(x^2+2x+3)
If 5<x<7,
the numerator is positive
the denominator is negative
So, no real value of x between 5 and 7 satisfies the condition that y > 0
That is, unless I have misinterpreted your somewhat garbled text.
If [(x-5)(x.x-2x+1)/(x-7)(x.x+2x+3)]is positive for real value of x, then prove that there is no value of x between 5 & 7.
3 answers
hi steve i need help so we have math it may seem easy to you and me but were doing old schoollike 24*32 and then,its like going on mulitiplication.
thanks@steve