Answer
$$\frac{(m^2+2-2)n^2-(m^2+2-4)n}{2}=\frac{m^2n^2-m^2n+2n}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(m^2+2-2)n^2-(m^2+2-4)n}{2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{m^2n^2+2n^2-2n^2-(1m^2n+2n-4n)}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{m^2n^2-(1m^2n-2n)}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{m^2n^2-m^2n+2n}{2}\end{aligned} $$ | |
| ① | $$ \left( \color{blue}{m^2+ \cancel{2} -\cancel{2}}\right) \cdot n^2 = m^2n^2+ \cancel{2n^2} -\cancel{2n^2} $$ |
| ② | $$ \left( \color{blue}{m^2+2-4}\right) \cdot n = m^2n+2n-4n $$ |
| ③ | Combine like terms: $$ m^2n^2+ \, \color{blue}{ \cancel{2n^2}} \, \, \color{blue}{ -\cancel{2n^2}} \, = m^2n^2 $$ |
| ④ | Combine like terms: $$ m^2n+ \color{blue}{2n} \color{blue}{-4n} = m^2n \color{blue}{-2n} $$ |
| ⑤ | Remove the parentheses by changing the sign of each term within them. $$ - \left( m^2n-2n \right) = -m^2n+2n $$ |
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