ĐKXĐ : \(x\ne3;5\)
Ta có : \(\dfrac{x+7}{x-3}+\dfrac{4x-2}{x-5}=5\)
\(\Leftrightarrow\dfrac{\left(x+7\right)\left(x-5\right)+\left(4x-2\right)\left(x-3\right)}{\left(x-3\right)\left(x-5\right)}=2\)
\(\Leftrightarrow x^2+7x-5x-35+4x^2-2x-12x+6=2\left(x-3\right)\left(x-5\right)\)
\(\Leftrightarrow5x^2-12x-29=2\left(x^2-3x-5x+15\right)\)
\(\Leftrightarrow5x^2-12x-29=2x^2-16x+30\)
\(\Leftrightarrow3x^2+4x-59=0\)
\(\Leftrightarrow3\left(x^2+\dfrac{4}{3}x-\dfrac{59}{3}\right)=0\)
\(\Leftrightarrow x^2+\dfrac{4}{3}x-\dfrac{59}{3}=0\)
\(\Leftrightarrow x^2+2x.\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{181}{9}=0\)
\(\Leftrightarrow\left(x+\dfrac{2}{3}\right)^2=\dfrac{181}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=\dfrac{\sqrt{181}}{3}\\x+\dfrac{2}{3}=\dfrac{-\sqrt{181}}{3}\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pm\sqrt{181}-2}{3}\)
Vậy ...
ĐKXĐ: \(x\ne3,x\ne5\)
Phương trình tương đương:
\(\dfrac{\left(x-5\right)\left(x+7\right)+\left(x-3\right)\left(4x-2\right)}{\left(x-3\right)\left(x-5\right)}=5\\ \Leftrightarrow\dfrac{5x^2-12x-29}{\left(x-3\right)\left(x-5\right)}=5\\ \Leftrightarrow5x^2-12x-29=5\left(x-3\right)\left(x-5\right)\\ \Leftrightarrow5x^2-12x-29=5x^2-25x-15x+75\\ \Leftrightarrow-12x-29=-25x-15x+75\\ \Leftrightarrow-12x-29=-40x+75\\ \Leftrightarrow-12x+40x=75+29\\ \Leftrightarrow28x=104\\ \Leftrightarrow x=\dfrac{26}{7}\left(TM\right)\)
Vậy \(S=\left\{\dfrac{26}{7}\right\}\)
