Question

a)\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)

b)\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)

c)\(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}=\dfrac{5-x}{2x^2+10x}\)

Giúp nha mn

Answer 1

a, \(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)

\(\Leftrightarrow\dfrac{12}{x^2-4}-\left(\dfrac{x+1}{x-2}-\dfrac{x+7}{x+2}\right)=0\)

\(\Leftrightarrow\dfrac{12}{x^2-4}-\left[\dfrac{\left(x+1\right)\left(x+2\right)-\left(x-2\right)\left(x+7\right)}{\left(x-2\right)\left(x+2\right)}\right]=0\)

\(\Leftrightarrow\dfrac{12}{x^2-4}-\left[\dfrac{x^2+3x+2-x^2-5x+14}{x^2-4}\right]=0\)

\(\Leftrightarrow\dfrac{12}{x^2-4}-\left(\dfrac{14-2x}{x^2-4}\right)=0\)

\(\Leftrightarrow12=14-2x\)

\(\Leftrightarrow x=1\)

Vậy x = 1