Stabilim conditiile de existenta a ecuatiei:
x + m ≠ 0 ⇒x ≠ -m; x - m ≠ 0 ⇒ x ≠ m
Deci, x ≠ ± m
[tex]\it...\Longleftrightarrow \dfrac{5}{m+x} -\dfrac{1}{m-x}-\dfrac{10m}{m^2-x^2} =2
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\Longleftrightarrow \dfrac{5m-5x-m-x-10m}{(m-x)(m+x)} =2
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\Longleftrightarrow \dfrac{-6m-6x}{(m-x)(m+x)} =2 \Leftrightarrow \dfrac{-6(m+x)}{(m-x)(m+x)} =2
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\Longleftrightarrow \dfrac{-6}{m-x}=2|_{:2} \Longleftrightarrow \dfrac{-3}{m-x} =1 \Longleftrightarrow -3=m-x \Longleftrightarrow x=m+3[/tex]
