I have been presented with the question, express the following as a single fraction:
$$\frac{1}{(x+1)(x+2)}-\frac{2}{x+1}+\frac{3}{x+2}$$
The answer in my worksheet is giving a different value to the calculated answer on a well known online calculator (Symbolab)
The answer from my worksheet is :
$$\frac{x}{(x+1)(x+2)}$$
The answer on Symbolab is :
$$\frac{x^2-2x+4}{(x+2)(x+1)(x-2)}$$
I am not sure which one is correct therefore not sure which one to work towards, and I do not know how to solve to attain either. Any guidance with the methodology of solving this would be very well received.
I'm assuming you intended:
$$ \frac{1}{(x+1)(x-2)} - \frac{2}{x+1} + \frac{3}{x +2} $$
In this case the first fraction is "missing" $(x+2)$, the second is "missing" $(x-2)(x+2)$, and the third is "missing" $(x+1)(x-2)$, which means you have:
\begin{align} \frac{(x+2) - 2(x-2)(x+2) + 3(x+1)(x-2)}{(x+1)(x-2)(x+2)} =&\ \frac{x+2 - 2(x^2-4) + 3(x^2-x-2)}{(x+1)(x-2)(x+2)} \\ =&\ \frac{x+2 -2x^2+8 + 3x^2-3x-6}{(x+1)(x-2)(x+2)} \\ =&\ \frac{x^2-2x+4}{(x+1)(x-2)(x+2)} \end{align}
Which is exactly what "Symbolab" gave you...because it's correct. It's always possible that the final numerator can be factored and thus cancel with part of the denominator, but not in this case since the determinant is $b^2-4ac = 4 - 16 = -12$ which, not only isn't a perfect square, doesn't even have real roots.
We can work backward from your supposed answer and find it's partial fraction decomposition:
\begin{align} \frac{x}{(x+1)(x+2)} =\ \frac{A}{x+1} + \frac{B}{x+2}\\ =&\ \frac{A(x+2) + B(x+1)}{(x+1)(x+2)} \\ =&\ \frac{(A +B)x + (2A+B)}{(x+1)(x+2)} \end{align}
This would lead to:
$$ A+B = 1 \rightarrow A = 1-B \\ 2A+B = 0 \rightarrow 2(1-B) + B = 0 \rightarrow 2 - B = 0 \rightarrow B = 2 \rightarrow A = -1 $$
So finally:
$$ \frac{x}{(x+1)(x+2)} = \frac{2}{x+2} - \frac{1}{x+1} $$
Edit (for comment):
To Ted Shifrin's point, if you take this amount (the answer your work sheet gave) and subtract it from your expression ignoring the $\frac{...}{(x+1)(x-2)}$ part, you get:
$$ \left(\zeta - \frac{2}{x + 1} + \frac{3}{x+2}\right) - \left(\frac{2}{x+2} - \frac{1}{x+1}\right) = 0 \\ \zeta -\frac{1}{x+1} + \frac{1}{x+2} = 0 \\ \zeta = \frac{1}{x+1} - \frac{1}{x+2} \\ \zeta = \frac{(x+2)-(x+1)}{(x+1)(x+2)} \\ \zeta = \frac{1}{(x+1)(x+2)} $$
Therefore, it's true that:
$$ \frac{1}{(x+1)(x+2)} - \frac{2}{x+1} + \frac{3}{x+2} = \frac{2}{x+2} - \frac{1}{x+1} = \frac{x}{(x+1)(x+2)} $$