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Now, we have to find the value of \[-\left( \dfrac{2}{3}\times \dfrac{3}{5} \right)+\left( \dfrac{5}{2} \right)-\left( \dfrac{3}{5}\times \dfrac{1}{6} \right)\].

Now, we know that according to the BODMAS Rule for solving the equation having different operations. The Rule BODMAS is a short form of Brackets, of, Division, Multiplication, Addition and Subtraction. This rule explains the order of operations to solve an expression. This rule gives the preference to the operation Brackets, of, Division, Multiplication, Addition and Subtraction respectively to solve the problem correctly.

Now, we have expression,

\[-\left( \dfrac{2}{3}\times \dfrac{3}{5} \right)+\left( \dfrac{5}{2} \right)-\left( \dfrac{3}{5}\times \dfrac{1}{6} \right)\]

So, as we have brackets here, we will first perform the operations inside the bracket according to the BODMAS, then we will perform multiplication of the terms inside the brackets as multiplication has the next highest preference. So, we will multiply terms in the expression.

So, first we will consider \[\dfrac{2}{3}\times \dfrac{3}{5}\], here we can cancel the like terms, so we will get $\dfrac{2}{5}$. Next, we will consider \[\dfrac{3}{5}\times \dfrac{1}{6}\], so on multiplying, we get $\dfrac{3}{30}=\dfrac{1}{10}$. Thus, after performing the multiplication of the terms, we can write our expression as,

\[-\left( \dfrac{2}{5} \right)+\left( \dfrac{5}{2} \right)-\left( \dfrac{1}{10} \right)\]

Now, addition has higher preference than subtraction. So, we will add \[-\dfrac{2}{5}\ and\ \dfrac{5}{2}\] by taking LCM. So, we know that LCM of 5 and 2 is 10. So, we can write,

\[\left( -\dfrac{2}{5}+\dfrac{5}{2} \right)=\dfrac{\left( -2\times 2 \right)+\left( 5\times 5 \right)}{10}=\dfrac{-4+25}{10}=\dfrac{21}{10}\]. Therefore, we can write our expression as,

$=\dfrac{21}{10}-\dfrac{1}{10}$

Now, we will perform the subtraction of the terms $\dfrac{21}{10}and\dfrac{1}{10}$. So, we get,

$\begin{align}

& =\dfrac{21-1}{10} \\

& =\dfrac{20}{10} \\

& =2 \\

\end{align}$