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**Hint:**This type of problem is based on the concept of division of polynomials. First, we need to place the polynomials in the long division table. Then, we should find a term which on multiplying with x, we get \[{{x}^{2}}\] as the first term. And as usual multiply x with x-2 and subtract the product with the polynomial \[{{x}^{2}}-5x+6\]. Then, find a term which on multiplying with x-2, we get -3x+6. Subtract the final term with the polynomial \[{{x}^{2}}-5x+6\]. Continue these steps again and again until we get 0 in the remainder. Once we get 0 in the remainder, the quotient is the required answer.

**Complete step by step answer:**

According to the question, we are asked to divide \[\dfrac{{{x}^{2}}-5x+6}{x-2}\].

We have been given the polynomials are \[{{x}^{2}}-5x+6\] and \[x-2\].

Let us represent the polynomials in the division table.

First, we need to find a term which on multiplying with x, we get \[{{x}^{2}}\] so that we can cancel \[{{x}^{2}}\].

We find that \[x\times x={{x}^{2}}\]. Then, we have to multiply x with x-2.

Therefore,

\[x-2\overset{x}{\overline{\left){\begin{align}

& {{x}^{2}}-5x+6 \\

& {{x}^{2}}-2x \\

\end{align}}\right.}}\]

Then, subtract the two polynomials \[{{x}^{2}}-5x+6\] and \[{{x}^{2}}-2x\].

On grouping the similar terms and subtracting, we get

\[{{x}^{2}}-5x+6-\left( {{x}^{2}}-2x \right)={{x}^{2}}-5x+6-{{x}^{2}}+2x\]

On further simplification, we get

\[\Rightarrow {{x}^{2}}-5x+6-\left( {{x}^{2}}-2x \right)=x\left( 2-5 \right)+6\]

\[\Rightarrow {{x}^{2}}-5x+6-\left( {{x}^{2}}-2x \right)=-3x+6\]

Therefore,

\[x-2\overset{x}{\overline{\left){\begin{align}

& {{x}^{2}}-5x+6 \\

& \dfrac{{{x}^{2}}-2x}{-3x+6} \\

\end{align}}\right.}}\]

Now, we need to find a term which on multiplying with (x-2), we get -3x+6.

We know that -3(x-2)=-3x+6.

Therefore, we get

\[x-2\overset{x-3}{\overline{\left){\begin{align}

& {{x}^{2}}-5x+6 \\

& \dfrac{{{x}^{2}}-2x}{\begin{align}

& -3x+6 \\

& -3x+6 \\

\end{align}} \\

\end{align}}\right.}}\]

Now, we need to subtract -3x+6 from -3x+6.

We know that the same polynomials on subtracting cancel out.

Therefore, we get

\[x-2\overset{x-3}{\overline{\left){\begin{align}

& {{x}^{2}}-5x+6 \\

& \dfrac{{{x}^{2}}-2x}{\begin{align}

& -3x+6 \\

& \dfrac{-3x+6}{0} \\

\end{align}} \\

\end{align}}\right.}}\]

Since the remainder is zero, we can stop the division.

\[\therefore \dfrac{{{x}^{2}}-5x+6}{x-2}=x-3\]

**Hence, the value of \[\dfrac{{{x}^{2}}-5x+6}{x-2}\] is x-3.**

**Note:**we can check whether the obtained answer is correct or not.

Multiply the answer with the divisor that is x-2. If they get the final product the same as the dividend, that is \[{{x}^{2}}-5x+6\], then the final answer obtained is correct.

Let us check.

We need to find \[\left( x-2 \right)\left( x-3 \right)\].

\[\Rightarrow \left( x-2 \right)\left( x-3 \right)=x\times x-2x-3x-3\times -2\]

Let us now simplify the above equation.

\[\Rightarrow \left( x-2 \right)\left( x-3 \right)={{x}^{2}}-5x+6\]

We have got the dividend as the product of x-2 and x-3.

Therefore, the answer obtained is verified.

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