$$\dfrac{\dfrac{2}{x}-4x}{2x+\dfrac{5}{x}}=\dfrac{-4x^2+2}{2x^2+5}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\frac{2}{x}-4x}{2x+\frac{5}{x}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\frac{-4x^2+2}{x}}{\frac{2x^2+5}{x}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-4x^2+2}{2x^2+5}\end{aligned} $$ | |
| ① | Subtract $4x$ from $ \dfrac{2}{x} $ to get $ \dfrac{ \color{purple}{ -4x^2+2 } }{ x }$. Step 1: Write $ 4x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: To subtract raitonal expressions, both fractions must have the same denominator. |
| ② | Add $2x$ and $ \dfrac{5}{x} $ to get $ \dfrac{ \color{purple}{ 2x^2+5 } }{ x }$. Step 1: Write $ 2x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: To add raitonal expressions, both fractions must have the same denominator. |
| ③ | Divide $ \dfrac{-4x^2+2}{x} $ by $ \dfrac{2x^2+5}{x} $ to get $ \dfrac{ -4x^2+2 }{ 2x^2+5 } $. Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction. Step 2: Cancel $ \color{red}{ x } $ in first and second fraction. Step 3: Multiply numerators and denominators. Step 4: Simplify numerator and denominator. $$ \begin{aligned} \frac{ \frac{-4x^2+2}{x} }{ \frac{\color{blue}{2x^2+5}}{\color{blue}{x}} } & \xlongequal{\text{Step 1}} \frac{-4x^2+2}{x} \cdot \frac{\color{blue}{x}}{\color{blue}{2x^2+5}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{-4x^2+2}{\color{red}{1}} \cdot \frac{\color{red}{1}}{2x^2+5} \xlongequal{\text{Step 3}} \frac{ \left( -4x^2+2 \right) \cdot 1 }{ 1 \cdot \left( 2x^2+5 \right) } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ -4x^2+2 }{ 2x^2+5 } \end{aligned} $$ |