`3/x + 9 -3/x =3/4` `(x≠0)`
suy ra
`12+36x-12=3x`
`<=> 36x-3x=-12+12`
`<=> 33x=0`
`<=> x=0(ktmđk)`
\(\dfrac{3}{x}+9-\dfrac{3}{x}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3+x.9-3}{x}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3+9x-3}{x}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{9x}{x}=\dfrac{3}{4}\)
\(\Leftrightarrow9=\dfrac{3}{4}\)
Vậy phương trình vô nghiệm
\(a,\dfrac{3}{x+9}-\dfrac{3}{x}=\dfrac{3}{4}\left(ĐKXĐ:x\ne-9\right)\)
\(\Leftrightarrow\dfrac{12x}{4x\left(x+9\right)}-\dfrac{12x\left(x+9\right)}{4x\left(x+9\right)}-\dfrac{3x\left(x+9\right)}{4x\left(x+9\right)}=0\)
\(\Leftrightarrow\dfrac{12x}{4x\left(x+9\right)}-\dfrac{12x^2+108x}{4x\left(x+9\right)}-\dfrac{3x^2+27x}{4x\left(x+9\right)}=0\)
\(\Leftrightarrow\dfrac{12x-12x^2-108x-3x^2-27x}{4x\left(x+9\right)}=0\)
\(\Rightarrow-15x^2-123x=0\)
\(\Leftrightarrow-3x\left(5x+41\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-3x=0\\5x+41=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{-41}{5}\left(tm\right)\end{matrix}\right.\)
