\(\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{x^2-13x+40}-1\)
điều kiện: \(x\ne5;8\)
\(\frac{6\left(x-8\right)+2\left(x-5\right)}{\left(x-5\right)\left(x-8\right)}-\frac{18}{x^2-13x+40}+1=0\)
\(\frac{6x-48+2x-10}{\left(x-5\right)\left(x-8\right)}-\frac{18}{x^2-8x-5x+40}+1=0\)
\(\frac{8x-58}{\left(x-5\right)\left(x-8\right)}-\frac{18}{x\left(x-8\right)-5\left(x-8\right)}+1=0\)
\(\frac{8x-58}{\left(x-5\right)\left(x-8\right)}-\frac{18}{\left(x-5\right)\left(x-8\right)}+\frac{\left(x-5\right)\left(x-8\right)}{\left(x-5\right)\left(x-8\right)}=0\)
\(\frac{8x-58-18+x^2-13x+40}{\left(x-5\right)\left(x-8\right)}=0\)
\(\frac{x^2-5x-36}{\left(x-5\right)\left(x-8\right)}=0\)
=> \(x^2-5x-36=0\)
\(x^2+4x-9x-36=0\)
\(x\left(x+4\right)-9\left(x+4\right)=0\)
\(\left(x-9\right)\left(x+4\right)=0\)
Vậy x - 9 = 0 hoặc x + 4 = 0
hay x = 9 (thỏa mãn điều kiện) hoặc x = -4 (thỏa mãn điều kiện)
vậy...
\(\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{x^2-13x+40}-1\)
ĐKXĐ: \(x\ne5,8\)
\(\Leftrightarrow\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{\left(x-5\right)\left(x-8\right)}-1\)
\(\Leftrightarrow6\left(x-8\right)+2\left(x-5\right)=18-\left(x-5\right)\left(x-8\right)\)
\(\Leftrightarrow8x-58=-22-x^2+13x\)
\(\Leftrightarrow8x-58+22+x^2-13x=0\)
\(\Leftrightarrow-5x-36+x^2=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-9=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=-4\end{cases}}\)
Vậy: phương trình có tập nghiệm là: S = {9; -4}
