$$n\dfrac{(m^2+2-2)n-(m^2+2-4)}{2}-\dfrac{((m+1)n-1)((m+1)n-1+1)}{2}=\dfrac{-m^2n-2mn^2+mn-n^2+3n}{2}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}n\frac{(m^2+2-2)n-(m^2+2-4)}{2}-\frac{((m+1)n-1)((m+1)n-1+1)}{2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}n\frac{m^2n+2n-2n-(1m^2-2)}{2}-\frac{(1mn+n-1)(1mn+n-1+1)}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}n\frac{m^2n-(1m^2-2)}{2}-\frac{(1mn+n-1)(1mn+n)}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}n\frac{m^2n-m^2+2}{2}-\frac{m^2n^2+mn^2+mn^2+n^2-mn-n}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} } }}}\frac{m^2n^2-m^2n+2n}{2}-\frac{m^2n^2+2mn^2-mn+n^2-n}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle11}{\textcircled {11}} } }}}\frac{-m^2n-2mn^2+mn-n^2+3n}{2}\end{aligned} $$ | |
| ① | $$ \left( \color{blue}{m^2+ \cancel{2} -\cancel{2}}\right) \cdot n = m^2n+ \cancel{2n} -\cancel{2n} $$ |
| ② | Combine like terms: $$ m^2+ \color{blue}{2} \color{blue}{-4} = m^2 \color{blue}{-2} $$ |
| ③ | $$ \left( \color{blue}{m+1}\right) \cdot n = mn+n $$ |
| ④ | $$ \left( \color{blue}{m+1}\right) \cdot n = mn+n $$ |
| ⑤ | Combine like terms: $$ m^2n+ \, \color{blue}{ \cancel{2n}} \, \, \color{blue}{ -\cancel{2n}} \, = m^2n $$ |
| ⑥ | Combine like terms: $$ mn+n \, \color{blue}{ -\cancel{1}} \,+ \, \color{blue}{ \cancel{1}} \, = mn+n $$ |
| ⑦ | Remove the parentheses by changing the sign of each term within them. $$ - \left( m^2-2 \right) = -m^2+2 $$ |
| ⑧ | Multiply each term of $ \left( \color{blue}{mn+n-1}\right) $ by each term in $ \left( mn+n\right) $. $$ \left( \color{blue}{mn+n-1}\right) \cdot \left( mn+n\right) = m^2n^2+mn^2+mn^2+n^2-mn-n $$ |
| ⑨ | Multiply $n$ by $ \dfrac{m^2n-m^2+2}{2} $ to get $ \dfrac{ m^2n^2-m^2n+2n }{ 2 } $. Step 1: Write $ n $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} n \cdot \frac{m^2n-m^2+2}{2} & \xlongequal{\text{Step 1}} \frac{n}{\color{red}{1}} \cdot \frac{m^2n-m^2+2}{2} \xlongequal{\text{Step 2}} \frac{ n \cdot \left( m^2n-m^2+2 \right) }{ 1 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ m^2n^2-m^2n+2n }{ 2 } \end{aligned} $$ |
| ⑩ | Combine like terms: $$ m^2n^2+ \color{blue}{mn^2} + \color{blue}{mn^2} +n^2-mn-n = m^2n^2+ \color{blue}{2mn^2} -mn+n^2-n $$ |
| ⑪ | Subtract $ \dfrac{m^2n^2+2mn^2-mn+n^2-n}{2} $ from $ \dfrac{m^2n^2-m^2n+2n}{2} $ to get $ \dfrac{-m^2n-2mn^2+mn-n^2+3n}{2} $. To subtract expressions with the same denominators, we subtract the numerators and write the result over the common denominator. $$ \begin{aligned} \frac{m^2n^2-m^2n+2n}{2} - \frac{m^2n^2+2mn^2-mn+n^2-n}{2} & = \frac{m^2n^2-m^2n+2n}{\color{blue}{2}} - \frac{m^2n^2+2mn^2-mn+n^2-n}{\color{blue}{2}} = \\[1ex] &=\frac{ m^2n^2-m^2n+2n - \left( m^2n^2+2mn^2-mn+n^2-n \right) }{ \color{blue}{ 2 }}= \frac{-m^2n-2mn^2+mn-n^2+3n}{2} \end{aligned} $$ |