\(\dfrac{x+3}{x-3}+\dfrac{3-x}{x+3}=\dfrac{36}{x^2-9}\left(x\ne3;x\ne-3\right)\)
\(< =>\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{x-3}{x+3}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\\ < =>\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
suy ra
\(x^2+6x+9-\left(x^2-6x+9\right)=36\\ < =>x^2+6x+9-x^2+6x-9=36\)
\(< =>x^2-x^2+6x+6x+9-9=36\)
\(< =>12x=36\\ < =>x=3\left(ktmđk\right)\)
Vậy phương trình vô nghiệm
\(\dfrac{x+3}{x-3}+\dfrac{3-x}{x+3}=\dfrac{36}{x^2-9}\)
\(\Leftrightarrow\dfrac{x+3}{x-3}+\dfrac{3-x}{x+3}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
Ta có : \(\dfrac{x+3}{x-3}+\dfrac{3-x}{x+3}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(3-x\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
`=> x^2 + 6x + 9 + 3x - 9 -x^2 + 3x -36=0`
`<=> 12x -36=0`
`<=> 12x=36`
`<=> x= 3(l)`
vậy pt vô nghiệm.
\(\dfrac{x+3}{x-3}+\dfrac{3-x}{x+3}=\dfrac{36}{x^2-9}\)
\(ĐKXĐ:x\ne\mp3\)
\(\dfrac{x+3}{x-3}+\dfrac{3-x}{x+3}=\dfrac{36}{x^2+3}\)
\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(3-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{36}{\left(x+3\right)\left(x-3\right)}\)
\(\Rightarrow x^2+3x+3x+9+3x-9-x^2+3x=36\)
\(\Leftrightarrow x^2+3x+3x+3x-x^2+3x=36-9+9\)
\(\Leftrightarrow12x=36\)
\(\Leftrightarrow x=3\left(ktm\right)\)
Vậy phương trình vô nghiệm: S={∅}
