Question

\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{10}{x^2-4}+1\)

Giải ptrình

Answer 1

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=10+x^2-4\)

\(\Leftrightarrow x^2+3x+2-5x+10-10-x^2+4=0\)

=>-2x+6=0

hay x=3(nhận)

Answer 2

đk : x khác 2 ; -2 

<=> x^2 + 3x + 2 - 5x + 10 = 10 + x^2 - 4 

<=> x^2 - 2x + 12 = x^2 + 6 

<=> -2x + 6 =0 <=> x = 3 (tm)

Answer 3

⇔ \(\dfrac{\left(x+1\right)\left(x+2\right)-5\left(x-2\right)}{x^2-4}=\dfrac{10+x^2-4}{x^2-4}\) (đk: x≠ +-2)

⇔ \(\dfrac{x^2+2x+x+2-5x+10-x^2-6}{x^2-4}=0\)

⇒ \(-2x+6=0\)

⇒ \(x=3\left(tm\right)\)