560. Решаи муодиларо ёбед: a) 15 x-2 14 =-+1; x б) 2 x-3 14 +-=3; x B) 1 x-3 1 15 +-=-: x 4 г) 1 x 1 x+3 3 +=--- 20 д) 40 x-20 40 =1; X e) 120 120 =1; X x+4 ж) 180 X 180 -1= x+6' 3) 2 x-1 4 +-=4. X

Решим уравнения: a) \[\frac{15}{x-2} = \frac{14}{x} + 1\] Приведем к общему знаменателю \(x(x-2)\): \[\frac{15x}{x(x-2)} = \frac{14(x-2)}{x(x-2)} + \frac{x(x-2)}{x(x-2)}\] \[15x = 14x - 28 + x^2 - 2x\] \[x^2 + 14x - 2x - 15x - 28 = 0\] \[x^2 - 3x - 28 = 0\] \( D = (-3)^2 - 4 \cdot 1 \cdot (-28) = 9 + 112 = 121\) \( x_1 = \frac{3 + 11}{2} = 7\) \( x_2 = \frac{3 - 11}{2} = -4\) б) \[ \frac{2}{x-3} + \frac{14}{x} = 3\] Приведем к общему знаменателю \(x(x-3)\): \[\frac{2x}{x(x-3)} + \frac{14(x-3)}{x(x-3)} = \frac{3x(x-3)}{x(x-3)}\] \( 2x + 14x - 42 = 3x^2 - 9x \) \( 3x^2 - 9x - 16x + 42 = 0 \) \( 3x^2 - 25x + 42 = 0 \) \( D = (-25)^2 - 4 \cdot 3 \cdot 42 = 625 - 504 = 121 \) \( x_1 = \frac{25 + 11}{6} = 6\) \( x_2 = \frac{25 - 11}{6} = \frac{7}{3}\) в) \[ \frac{1}{x-3} + \frac{1}{x} = -\frac{5}{4}\] Приведем к общему знаменателю \(4x(x-3)\): \[\frac{4x}{4x(x-3)} + \frac{4(x-3)}{4x(x-3)} = -\frac{5x(x-3)}{4x(x-3)}\] \( 4x + 4x - 12 = -5x^2 + 15x \) \( 5x^2 + 8x - 15x - 12 = 0 \) \( 5x^2 - 7x - 12 = 0 \) \( D = (-7)^2 - 4 \cdot 5 \cdot (-12) = 49 + 240 = 289 \) \( x_1 = \frac{7 + 17}{10} = \frac{12}{5} \) \( x_2 = \frac{7 - 17}{10} = -1 \) г) \[ \frac{1}{x} + \frac{1}{x+3} = -\frac{3}{20}\] Приведем к общему знаменателю \(20x(x+3)\): \[\frac{20(x+3)}{20x(x+3)} + \frac{20x}{20x(x+3)} = -\frac{3x(x+3)}{20x(x+3)}\] \[ 20x + 60 + 20x = -3x^2 - 9x \] \[ 3x^2 + 40x + 9x + 60 = 0 \] \[ 3x^2 + 49x + 60 = 0 \] \( D = 49^2 - 4 \cdot 3 \cdot 60 = 2401 - 720 = 1681\) \( x_1 = \frac{-49 + 41}{6} = -\frac{4}{3} \) \( x_2 = \frac{-49 - 41}{6} = -15 \) д) \[ \frac{40}{x-20} - \frac{40}{x} = 1\] Приведем к общему знаменателю \(x(x-20)\): \[\frac{40x}{x(x-20)} - \frac{40(x-20)}{x(x-20)} = \frac{x(x-20)}{x(x-20)}\] \[ 40x - 40x + 800 = x^2 - 20x \] \[ x^2 - 20x - 800 = 0 \] \( D = (-20)^2 - 4 \cdot 1 \cdot (-800) = 400 + 3200 = 3600 \) \( x_1 = \frac{20 + 60}{2} = 40 \) \( x_2 = \frac{20 - 60}{2} = -20 \) e) \[ \frac{120}{x} - \frac{120}{x+4} = 1\] Приведем к общему знаменателю \(x(x+4)\): \[\frac{120(x+4)}{x(x+4)} - \frac{120x}{x(x+4)} = \frac{x(x+4)}{x(x+4)}\] \( 120x + 480 - 120x = x^2 + 4x \) \( x^2 + 4x - 480 = 0 \) \( D = 4^2 - 4 \cdot 1 \cdot (-480) = 16 + 1920 = 1936 \) \( x_1 = \frac{-4 + 44}{2} = 20 \) \( x_2 = \frac{-4 - 44}{2} = -24 \) ж) \[ \frac{180}{x} - 1 = \frac{180}{x+6}\] Приведем к общему знаменателю \(x(x+6)\): \[\frac{180(x+6)}{x(x+6)} - \frac{x(x+6)}{x(x+6)} = \frac{180x}{x(x+6)}\] \( 180x + 1080 - x^2 - 6x = 180x \) \( x^2 + 6x - 1080 = 0 \) \( D = 6^2 - 4 \cdot 1 \cdot (-1080) = 36 + 4320 = 4356 \) \( x_1 = \frac{-6 + 66}{2} = 30 \) \( x_2 = \frac{-6 - 66}{2} = -36 \) 3) \[ \frac{2}{x-1} + \frac{4}{x} = 4\] Приведем к общему знаменателю \(x(x-1)\): \[\frac{2x}{x(x-1)} + \frac{4(x-1)}{x(x-1)} = \frac{4x(x-1)}{x(x-1)}\] \( 2x + 4x - 4 = 4x^2 - 4x \) \( 4x^2 - 4x - 6x + 4 = 0 \) \( 4x^2 - 10x + 4 = 0 \) \( 2x^2 - 5x + 2 = 0 \) \( D = (-5)^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9 \) \( x_1 = \frac{5 + 3}{4} = 2 \) \( x_2 = \frac{5 - 3}{4} = \frac{1}{2} \)

Ответ: a) x = 7, x = -4; б) x = 6, x = 7/3; в) x = 12/5, x = -1; г) x = -4/3, x = -15; д) x = 40, x = -20; e) x = 20, x = -24; ж) x = 30, x = -36; 3) x = 2, x = 1/2

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