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.

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.

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

To prove that there is no value of x between 5 and 7 for which the expression

[(x-5)(xx-2x+1)/(x-7)(xx+2x+3)]

is positive, we need to determine when the expression is positive or negative for values of x within that range.

First, let's simplify the expression:

(x-5)(xx-2x+1)/(x-7)(xx+2x+3)

= (x-5)(x^2 - 2x + 1) / (x-7)(x^2 + 2x + 3)

Before proceeding further, we need to identify any discontinuities or values that make the denominator zero and exclude them from our analysis. In this case, the denominator can be zero at x = 7 since (x-7) is in the denominator.

Now, let's determine when the numerator is positive or negative.

For the numerator (x-5)(x^2 - 2x + 1), we can see that it will always be positive or zero for any real value of x because it is a quadratic expression multiplied by a linear expression, and the square term (x^2) is always non-negative.

Next, let's examine the denominator (x-7)(x^2 + 2x + 3).

At x = 7, the denominator becomes zero, and we need to exclude this value from our analysis.

Now, let's consider the sign of the denominator for values of x between 5 and 7.

Plug in a value within that range, such as x = 6:

(6-7)(6^2 + 2(6) + 3) = (-1)(36 + 12 + 3) = (-1)(51) = -51.

Since the denominator evaluated to a negative value for x = 6, we can conclude that the expression is negative for every value of x between 5 and 7.

Hence, there is no value of x between 5 and 7 for which the given expression is positive.