\(\dfrac{3x-x}{x+2}+\dfrac{3x+x}{x-2}=\dfrac{1}{x^2-4}\)
\(\dfrac{\left(3x-x\right)\left(x-2\right)}{x^2-4}+\dfrac{\left(3+x\right)\left(x+2\right)}{x^2-4}=\dfrac{1}{x^2-4}\)
\(\dfrac{2x\left(x-2\right)+\left(3+x\right)\left(x+2\right)}{x^2-4}=\dfrac{1}{x^2-4}\)
\(2x^2-4x+x^2+5x+6=1\)
\(3x^2+x+6=1\)
\(3x^2+x+5=0\)
Tới đây tính delta thì ra vô nghiệm cho nhanh nhưng thôi.
\(2x^2+x^2+x+\dfrac{1}{4}+\dfrac{19}{4}=0\)
\(2x^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}=0\)
\(2x^2\ge0;\left(x+\dfrac{1}{2}\right)^2\ge0\)
\(\Rightarrow2x^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}>0\)
Vậy: Phương trình vô nghiệm
\(\dfrac{3x-x}{x+2}+\dfrac{3+x}{x-2}=\dfrac{1}{x^2-4}\)
or
\(3x-\dfrac{x}{x+2}+\dfrac{3+x}{x-2}=\dfrac{1}{x^2-4}\)
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