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6/(x^2-3x)=12/x+1/(x-3)

Sagot :

[tex]\\ \rm\Rrightarrow \dfrac{6}{x^2-3x}=\dfrac{12}{x}+\dfrac{1}{x-3}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{6}{x(x-3)}=\dfrac{12}{x}+\dfrac{1}{x-3}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{6}{x(x-3)}=\dfrac{12(x-3)+x}{x(x-3)}[/tex]

[tex]\\ \rm\Rrightarrow 6=12x-36+x[/tex]

[tex]\\ \rm\Rrightarrow 6=13x-36[/tex]

[tex]\\ \rm\Rrightarrow 13x=42[/tex]

[tex]\\ \rm\Rrightarrow x=\dfrac{42}{13}[/tex]

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