Question

giải pt
\(\dfrac{260}{x+2}-\dfrac{180}{x}=3\)

Answer 1

ĐKXĐ: x ≠ -2; x ≠ 0

pt <=> \(\dfrac{260x-180\left(x+2\right)}{x\left(x+2\right)}=3\Leftrightarrow260x-180x-360=3x^2+6x\Leftrightarrow3x^2+86x+360=0\Leftrightarrow3\left(x-\dfrac{20}{3}\right)\left(x-18\right)=0\Leftrightarrow\left[{}\begin{matrix}x=18\\x=\dfrac{20}{3}\end{matrix}\right.\left(TM\right)}\)

Answer 2

\(\Leftrightarrow260x-180\left(x+2\right)=3x\left(x+2\right)\)

=>3x^2+6x=80x-360

=>3x^2-74x+360=0

hay \(x\in\varnothing\)