Question

[Dennis]

Giải phương trình :

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)

Answer 1

[Phan Cả Phát]

Theo bài ra , ta có :

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)

ĐKXĐ : \(x\ne3,x\ne-3,x\ne-\frac{7}{2}\)

Quy đồng và khử mẫu phương trình ta đk :

\(13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow\left(x+3\right)\left(13+x-3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow\left(x+3\right)\left(x+10\right)=12x+42\)

\(\Leftrightarrow x^2+13x+30=12x+42\)

\(\Leftrightarrow x^2+13x-12x+30-42=0\)

\(\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow x^2-3x+4x-12=0\)

\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

Kết hợp với ĐKXĐ ta có : x = -4

Vậy \(S=\left\{-4\right\}\)

Chúc bạn học tốt =))

Answer 2

[Ken Tom Trần]

ĐKXĐ: x\(\ne\)3;-7/2;-3

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\Leftrightarrow\frac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\frac{\left(x-3\right)\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\frac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)

\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow13x+39+x^2-9=12x+42\\ \Leftrightarrow x^2+x=12\)

\(\Leftrightarrow x^2+x-12=0\Leftrightarrow x^2-3x+4x-12=0\\ \Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\Leftrightarrow\left[\begin{matrix}x-3=0\Rightarrow x=3\\x+4=0\Rightarrow x=-4\end{matrix}\right.\)

Nhận thấy x=3 không thỏa mãn ĐKXĐ nên pt có 1 nghiệm duy nhất là x=-4