ĐK: ` x \ne 10; x \ne 0`
`120/(x-10)-3/5=120/x`
`<=>120/(x-10)-120/x=3/5`
`<=>1/(x-10) - 1/x= 1/200`
`<=> (x-x+10)/(x(x-10)) = 1/200`
`<=> 10/(x(x-10))= 1/200`
`<=> x^2-10=2000`
`<=>` \(\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\)
Vậy `S={50;-40}`.
`120/(x-10)-3/5=120/x(x ne 0,x ne 10)`
`<=>40/(x-10)-1/5=40/x`
`<=>200x-x(x-10)=200(x-10)`
`<=>200x-200x+2000-x^2+10x=0`
`<=>x^2-10x-2000=0`
`Delta'=25+2000=2025`
`<=>x_1=50,x_2=-40`
Vậy `S={50,-40}`
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne10\\x\ne0\end{matrix}\right.\)
Ta có : \(\dfrac{120}{x-10}-\dfrac{3}{5}=\dfrac{120}{x}=\dfrac{600-3\left(x-10\right)}{5\left(x-10\right)}\)
\(\Leftrightarrow600\left(x-10\right)=600x-3x\left(x-10\right)\)
\(\Leftrightarrow600x-6000=600x-3x^2+30x\)
\(\Leftrightarrow3x^2-30x-6000=0\)
\(\Leftrightarrow\left(x-50\right)\left(x+40\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\) ( TM )
Vậy ...
