\(\dfrac{2\left(x^2+2\right)}{x^2-4}-\dfrac{x-1}{x+2}=\dfrac{x+1}{x-2}\)
(ĐKXĐ : x ≠ 2; x ≠ -2)
\(\Rightarrow2\left(x^2+2\right)-\left(x-1\right)\left(x-2\right)=\left(x+1\right)\left(x+2\right)\)
\(\Leftrightarrow2x^2+4-x^2-2x-x+2=x^2+2x+x+2\)
\(\Leftrightarrow0x=4\) (vô nghiệm)
Vậy : S = { ∅ }
ĐKXĐ:x khác +-2
\(\dfrac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{2x^2+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x^2-2x-x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x^2+2x+x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
khử mẫu:
=>2x2+2-x2-2x-x+2=x2+2x+x+2
=>2x2+4-x2-2x-x+2-x2-2x-x-2=0
=>0x+4=0
=>x vô nghiệm
đk : x khác 2 ; -2
\(2x^2+4-\left(x-1\right)\left(x-2\right)=\left(x+1\right)\left(x+2\right)\)
\(\Leftrightarrow2x^2+4-x^2+3x-2=x^2+3x+2\Leftrightarrow3x-2=3x+2\)
Vậy pt vô nghiệm