\(< =>\dfrac{13\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-9}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\dfrac{6\left(2x+7\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}\left(ĐK:x\ne\left\{-\dfrac{7}{2};3;-3\right\}\right)\\ =>13x+39+x^2-9=12x+42\\ < =>x^2+x-12=0\\ < =>\left(x+4\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x=-4\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\\ =>S=\left\{-4\right\}\)
\(ĐKXĐ:x\ne\dfrac{7}{2}\) và \(x\ne\pm3\)
mẫu chung : \(\left(2x+7\right)\left(x+3\right)\left(x-3\right)\)
Khử mẫu ta được :
\(13\left(x+3\right)+\left(x+3\right)\left(x-3\right)=6\left(2x+7\right)\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)
\(x=\left\{{}\begin{matrix}-4\\3\end{matrix}\right.\)
do \(x=3\) không thỏa mãn điều kiện thích hợp nên pt có nghiệm duy nhất là : \(-4\)
\(Vậy...\)
Tách x^2 - 9 ra thành x+3 nhân x-3
Mẫu có 2x+7, x-3,x+3, đưa về 1 vế quy đồng nhé!
\(\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}=\dfrac{6}{x^2-9}\left(đk:x\ne-\dfrac{7}{2};x\ne\pm3\right)\)
`<=>`\(\dfrac{13\left(x+3\right)+\left(x+3\right)\left(x-3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\dfrac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)
`=> 13x + 39 + x^2 -9 = 12x +42`
`<=> x^2 +13x -12 x +39-9-42 =0`
`<=> x^2 +x + -12 =0`
`<=> x^2 +4x -3x -12 =0`
`<=> (x-3)(x+4) =0`
`<=>[(x=-4);(x=3(ktm)):}`
`=> x=-4`
Vậy `S={-4}
