Answer
$ \frac{x(x+1)}{2x+5}$.
Work Step by Step
In order to simplify the given expression we first factor all numerators and denominators.
Factor $2x^2+x-1$.
Rewrite the middle term $x$ as $2x-1x$.
$\Rightarrow 2x^2+2x-1x-1$
Group the terms.
$\Rightarrow (2x^2+2x)+(-1x-1)$
Factor each group.
$\Rightarrow 2x(x+1)-1(x+1)$
Factor out $(x+1)$.
$\Rightarrow (x+1)(2x-1)$
Factor $2x^2-9x+4$.
Rewrite the middle term $-9x$ as $-8x-1x$.
$\Rightarrow 2x^2-8x-1x+4$
Group the terms.
$\Rightarrow (2x^2-8x)+(-1x+4)$
Factor each group.
$\Rightarrow 2x(x-4)-1(x-4)$
Factor out $(x-4)$.
$\Rightarrow (x-4)(2x-1)$
Factor $6x+15$.
Factor out $3$.
$\Rightarrow 3(2x+5)$
Factor $3x^2-12x$.
Factor out $3x$.
$\Rightarrow 3x(x-4)$
Back substitute all factors into the given expression.
$\Rightarrow \frac{2x^2+x-1}{2x^2-9x+4}\div\frac{6x+15}{3x^2-12x}=\frac{(x+1)(2x-1)}{(x-4)(2x-1)}\div\frac{3(2x+5)}{3x(x-4)}$
Invert the divisor and multiply.
$\Rightarrow \frac{(x+1)(2x-1)}{(x-4)(2x-1)}\times\frac{3x(x-4)}{3(2x+5)}$
Cancel common factors.
$\Rightarrow \frac{x(x+1)}{2x+5}$.
