Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{x}-\frac{1}{x+1}\frac{x}{x^2+2x+1}+\frac{1}{x+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{x}-\frac{x}{x^3+3x^2+3x+1}+\frac{1}{x+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^3+2x^2+3x+1}{x^4+3x^3+3x^2+x}+\frac{1}{x+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2x^3+4x^2+4x+1}{x^4+3x^3+3x^2+x}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{1}{x+1} $ by $ \dfrac{x}{x^2+2x+1} $ to get $ \dfrac{x}{x^3+3x^2+3x+1} $. Step 1: Multiply numerators and denominators. Step 2: Simplify numerator and denominator. $$ \begin{aligned} \frac{1}{x+1} \cdot \frac{x}{x^2+2x+1} & \xlongequal{\text{Step 1}} \frac{ 1 \cdot x }{ \left( x+1 \right) \cdot \left( x^2+2x+1 \right) } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ x }{ x^3+2x^2+x+x^2+2x+1 } = \frac{x}{x^3+3x^2+3x+1} \end{aligned} $$ |
| ② | Subtract $ \dfrac{x}{x^3+3x^2+3x+1} $ from $ \dfrac{1}{x} $ to get $ \dfrac{ \color{purple}{ x^3+2x^2+3x+1 } }{ x^4+3x^3+3x^2+x }$. To subtract raitonal expressions, both fractions must have the same denominator. |
| ③ | Add $ \dfrac{x^3+2x^2+3x+1}{x^4+3x^3+3x^2+x} $ and $ \dfrac{1}{x+1} $ to get $ \dfrac{ \color{purple}{ 2x^3+4x^2+4x+1 } }{ x^4+3x^3+3x^2+x }$. To add raitonal expressions, both fractions must have the same denominator. |