Question

Answer 1

b. \(\dfrac{x+1}{x-2}-\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)

<=> \(\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}-\dfrac{\left(x-1\right)\left(x-2\right)}{x^2-4}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)

=> (x + 1)(x + 2) - (x - 1)(x - 2) = 2(x² + 2)

<=> x² + 2x + x + 2 - x² + 2x + x - 2 = 2x² + 4

<=> x² - x² - 2x² + 2x + x + 2x + x + 2 - 2 - 4 = 0

<=> -2x² + 6x - 4 = 0

<=> -(-2x² + 6x - 4) = -0

<=> 2x² - 6x + 4 = 0

<=> 2x² - 4x - 2x + 4 = 0

<=> 2x(x - 2) - 2(x - 2) = 0

<=> (2x - 2)(x - 2) = 0

<=> \(\left[{}\begin{matrix}2x-2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Answer 2

c: Ta có: \(x^3+1=x\left(x+1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1-x\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)