Question

\(\dfrac{2}{x-1}\)+\(\dfrac{2x+3}{x^2+x+1}\)=\(\dfrac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)

Answer 1

\(\dfrac{2}{x-1}+\dfrac{2x+3}{x^2+x+1}=\dfrac{\left(2x+1\right)\left(2x-1\right)}{x^3-1}\)(đkxđ: x \(\ne\) 1)

\(\Leftrightarrow\) \(\dfrac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x-1\right)\left(2x+3\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4x^2-1}{x^3-1}\)

\(\Leftrightarrow\) \(\dfrac{2x^2+2x+2}{x^3-1}+\dfrac{2x^2+3x-2x-3}{x^3-1}=\dfrac{4x^2-1}{x^3-1}\)

\(\Leftrightarrow\) 2x² + 2x + 2 + 2x² + x - 3 = 4x² - 1

\(\Leftrightarrow\) 4x² + 3x - 1 = 4x² - 1

\(\Leftrightarrow\) 4x² - 4x² + 3x = 1 - 1

\(\Leftrightarrow\) 3x = 0

\(\Leftrightarrow\) x = 0(t/m điều kiện)