how would i go on about solving this? give me some steps to understand plz
There's nothing to 'solve' here, but we can simplify like so:
$\displaystyle \frac{x^2-1}{5x^2+5x}*\frac{15x^2}{3x-3}$
$\displaystyle =\frac{(x-1)(x+1)}{5x(x+1)}*\frac{3(5x^2)}{3(x-1)}$
Looks like we can do some cancelling here......
$\displaystyle =\frac{(x-1)}{5x}*\frac{5x^2}{(x-1)}$
and more cancelling.........
$\displaystyle \frac{1}{1}*\frac{x}{1}=x$
How 'bout that?
The other problems similar. You can do it. I know you can.
fOR THE DIVISION PROBLEMS,JUST FLIP THE FRACTION ON THE RIGHT AND MULTIPLY. JUST BECAUSE YOU'RE DEALING WITH MORE COMLICATED NUMERATORS AND DENOMINATORS, THEY'RE STILL JUST FRACTIONS.
Factorise whatever you can and try to cancel out common factors. Remember that $\displaystyle \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$ and $\displaystyle \frac{a}{b} \div \frac{c}{d} = \frac{ad}{bc}$
I'll go through the first one:
\frac{x^2-1}{5x^2+5x} \cdot \frac{15x^2}{3x-3}
x^2-1 is the difference of two squares: $\displaystyle (x^2-1) = (x-1)(x+1)$
5x^2+5x has a common factor of 5x: $\displaystyle 5x(x+1)$
Now we see that an (x+1) can cancel to give the LHS:
$\displaystyle \frac{x-1}{5x}$
15x^2 doesn't need factorising, it will cancel in due course
3x-3 has a common factor of 3: $\displaystyle 3x-3 = 3(x-1)$
On the RHS we can cancel a 3: $\displaystyle \frac{5x^2}{x-1}$
bring the two sides together:
$\displaystyle \frac{x-1}{5x} \times \frac{5x^2}{x-1}$
5x will cancel and so will x-1 which leaves a final answer of x