Please check for me.

7p - 7/p divided by 10p-10/5p^2

= 7p^2

6x - 6/x * 6x^2/8x-8

= 9x/2

I'm not sure how to answer this one.

x/2 divided by 9/x + 5

I think it's x(x + 5) / 18

Without grouping symbols () it is impossible to understand the problems.

The ^ means exponent and / is a fraction

(7p - 7/p) / (10p-10/5p^2)

= 7p^2

(6x - 6/x) * (6x^2/8x-8)

= 9x/2

I'm not sure how to answer this one.

(x/2) / (9/x + 5)

I think it's x(x + 5) / 18

For your first one:

(7p - 7/p) / (10p-10/5p^2)

Let's find common denominators and try to simplify from there.

Using p as a common denominator in the numerator, we have this:

(7p^2/p - 7/p) / [ 10(p - 1)/5p^2) ]

Going from there, we have:

[ (7p^2 - 7)/p ] / [ 2(p - 1)/p^2 ]

[ 7(p^2 - 1)/p ] / [ 2(p - 1)/p^2 ]

[ 7(p + 1)(p - 1)/p ] * [ p^2/2(p - 1) ]

Cancel out common terms in the numerator and denominator of the fractions.

When you do this, you will end up with:

7p(p + 1)/2

And that's as far as we can go on this one.

For your second problem:
(6x - 6/x) * (6x^2/8x-8)

(6x^2/x - 6/x) * [ 6x^2/8(x - 1) ]

(6x^2 - 6)/x * [ 6x^2/8(x - 1) ]

[ 6(x^2 - 1)/x ] * [ 6x^2/8(x - 1) ]

[ 6(x + 1)(x - 1)/x ] * [ 6x^2/8(x - 1) ]
Cancel out like terms and reduce where you can to end up with this:

9x(x + 1)/2

And that's as far as we can go on this one.

For your final problem:
(x/2) / [ 9/(x + 5) ]

Your answer is correct!

I hope this helps.

Based on the given expression, (x/2) / (9/(x + 5)), we can simplify it by multiplying the numerator and denominator by the reciprocal of the fraction in the denominator.

Step 1: To multiply a fraction by its reciprocal, we swap the numerator and denominator of the fraction. So the reciprocal of 9/(x + 5) is (x + 5)/9.

Step 2: Now we can rewrite the expression as (x/2) * [(x + 5)/9].

Step 3: Next, we can simplify the expression by multiplying the numerators and denominators. So (x * (x + 5))/(2 * 9) = (x(x + 5))/(18).

Therefore, the simplified expression is x(x + 5)/18.

Great job on getting the correct answer! If you have any more questions, feel free to ask.