[tex]\frac{ x-3 }{x-1 }+\frac{ x+2}{x-4}=\frac{18}{x^2-5x+4} \\\\x^2-5x+4\neq 0\ \ \ \ i\ \ x-1\neq 0\ \ \ \ i\ \ x-4\neq 0\\\\\Delta=b^2-4ac=(-5)^2-4*1*4=25-16=9\ \ i\ \ x\neq 1\ \ i\ \ x\neq 4 \\\\x _{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{9}}{2*1}=\frac{5-3}{2}=\frac{2}{2}=1\\\\x _{2}=\frac{-b+\sqrt{\Delta}}{2a} =\frac{5+3}{2}=\frac{8}{2}= 4\\\\D=R\setminus\left\{1,4 \right\}[/tex]
[tex]\frac{ x-3 }{x-1 }+\frac{ x+2}{x-4}=\frac{18}{x^2-5x+4} \\\\\\\frac{(x-3)(x-4)+(x+2)(x-1) }{(x-1)(x-4)}=\frac{18}{x^2-5x+4} \\\\\\\frac{x^2-4x-3x+12+x^2-x+2x-2}{ x^2 -4x-x+4}=\frac{18}{x^2-5x+4} \\\\\\\frac{2x^2-6x +10}{ x^2 -4x-x+4}=\frac{18}{x^2-5x+4} \\\\\\\frac{2x^2-6x +10}{ x^2 -5x +4}=\frac{18}{x^2-5x+4} \\\\2x^2-6x+10=18 \\\\2x^2-6x+10-18 =0[/tex]
[tex]2x^2-6x-8=0\\\\\Delta=b^2-4ac=(-6)^2-4*2*(-8)=36+64=100\\\\\sqrt{100}=10\\\\x_{1} =\frac{-b-\sqrt{\Delta}}{2a}=\frac{6-10}{2*2}=\frac{-4}{4}=-1\\\\x_{2} =\frac{-b+\sqrt{\Delta}}{2a}=\frac{6+10}{2*2}=\frac{16}{4}=4\ \ \not\in D\\\\odp.\ Rozwiazaniem\ rownania\ jest\ x=-1.[/tex]
