\(\dfrac{x-1}{x+2}+\dfrac{x+1}{x-2}=\dfrac{6}{4-x^2}\left(dkxd:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{x-1}{x+2}+\dfrac{x+1}{x-2}=-\dfrac{6}{x^2-4}\)
\(\Leftrightarrow\dfrac{x-1}{x+2}+\dfrac{x+1}{x-2}+\dfrac{6}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)+\left(x+1\right)\left(x+2\right)+6}{x^2-4}=0\)
\(\Leftrightarrow x^2-2x-x+2+x^2+2x+x+2+6=0\)
\(\Leftrightarrow2x^2+10=0\)
\(\Leftrightarrow2\left(x^2+5\right)=0\)
\(\Leftrightarrow x^2=-5\left(ktm\right)\)
Vậy \(S=\varnothing\)
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