[tex]\frac{2x + 3}{x^{2} -4x+4} - \frac{x - 1}{x^{2} - 2x} = \frac{5}{x}, x\neq 2, x\neq 0\\\\\frac{2x + 3}{x^{2} -4x+4} - \frac{x - 1}{x^{2} - 2x} - \frac{5}{x} = 0\\\\ \frac{2x + 3}{(x- 2)^{2} } - \frac{x - 1}{x(x-2)} - \frac{5}{x} = 0\\\\ \frac{x(2x + 3) - (x-2)(x-1) - 5(x-2)^{2}}{x(x-2)^{2}}=0\\\\ \frac{2x^{2} +3x-(x^{2} -x-2x+2)-5(x^{2} -4x+4)}{x(x^{2} -4x+4)}=0 \\\\\frac{2x^{2} +3x-(x^{2} -3x+2)-5x^{2} +20x-20}{x(x^{2} -4x+4)} = 0\\\\[/tex]

[tex]\frac{2x^{2} +3x-x^{2} +3x-2-5x^{2} +20x-20}{x(x^{2} -4x+4)} = 0\\\\\frac{x^{2} +6x-2-5x^{2} +20x-20}{x(x^{2} -4x+4)} = 0\\\\\frac{-4x^{2} +26x-22}{x(x^{2} -4x+4)}=0\\\\ -4x^{2} +26x-22 = 0\\\\2x^{2} -13x+11=0\\\\[/tex]

Далі квадратне рівняння:

[tex]D = b^{2} - 4ac =(-13)^{2} -4*2*11=81\\x_{1} = \frac{-b + \sqrt{D} }{2a} = \frac{-(-13) + 9}{2 * 2}= \frac{11}{2} \\\\x_{2} = \frac{-b - \sqrt{D} }{2a} = \frac{-(-13) - 9}{2 * 2} = 1\\\\x_{1} = \frac{11}{2}\\\\x_{2} = 1[/tex]

Ответ:

Объяснение:..........