Question

Giải phương trình sau:

\(\dfrac{2}{x^2+2x+1}-\dfrac{5}{x^2-2x+1}=\dfrac{3}{1-x^2}\)

Answer 1

\(\dfrac{2}{x^2+2x+1}-\dfrac{5}{x^2-2x+1}=\dfrac{3}{1-x^2}\)

⇔ \(\dfrac{2}{\left(x+1\right)^2}-\dfrac{5}{\left(x-1\right)^2}=\dfrac{-3}{\left(x+1\right)\left(x-1\right)}\) ( x # 1 ; x # -1)

⇔\(\dfrac{2\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}-\dfrac{5\left(x+1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}=\dfrac{-3\left(x^2-1\right)}{\left(x+1\right)^2\left(x-1\right)^2}\)

⇔ 2x² - 4x + 2 - 5x² - 10x - 5 = -3x² + 3

⇔ -3x² - 14x - 3 + 3x² - 3 = 0

⇔ - 14x - 6 = 0

⇔ -2( 7x + 3) = 0

⇔ 7x + 3 = 0

⇔ x = \(\dfrac{-3}{7}\) ( TM)

KL....