Question

Giải phương trình

\(\frac{6x+1}{x^2-7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)

Answer 1

\(\frac{6x+1}{x^2-7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\\ \Leftrightarrow\frac{6x+1}{x^2-7x+10}+\frac{5x-25}{x^2-7x+10}-\frac{3x-6}{x^2-7x+10}=0\\ \Leftrightarrow\frac{6x+1+5x-25-3x+6}{x^2-7x+10}=0\\ \Leftrightarrow\frac{8x-18}{x^2-7x+10}=0\\ \Rightarrow8x-18=0\\ \Rightarrow x=\frac{18}{8}=\frac{9}{4}\)

Answer 2

\(\frac{6x+1}{x^2-7x+10}\)+\(\frac{5}{x-2}\)=\(\frac{3}{x-5}\) (ĐKXĐ:x≠2;x≠5)

⇔\(\frac{6x+1}{\left(x-2\right)\left(x-5\right)}\)+\(\frac{5\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}\)=\(\frac{3\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}\)

⇒6x+1+5x-25=3x-6

⇔6x+5x-3x=-6-1+25

⇔8x=18

⇔x=\(\frac{9}{4}\) (TMĐK)

Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{9}{4}\)}