Question
Answer
$$1+\dfrac{0.02}{(x-1)(x+0.7)}=1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}1+\frac{0.02}{(x-1)(x+0.7)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}1+\frac{0.02}{x^2+0x-x+0} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}1+\frac{0.02}{x^2-x} \xlongequal{ } \\[1 em] & \xlongequal{ }1+0 \xlongequal{ } \\[1 em] & \xlongequal{ }1\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x-1}\right) $ by each term in $ \left( x0\right) $. $$ \left( \color{blue}{x-1}\right) \cdot \left( x0\right) = x^20x-x0 $$ |
| ② | Combine like terms: $$ x^2 \color{blue}{0x} \color{blue}{-x} 0 = x^2 \color{blue}{-x} $$ |
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