Решите уравнение: 1) \(\frac{x^2-9x}{x+3} = \frac{36}{x+3};\) 2) \(\frac{x^2+x}{x^2-25} = \frac{45-3x}{x^2-25};\) 3) \(\frac{5x-8}{x-1} = \frac{14x+12}{3x+5};\) 4) \(\frac{1}{x-4} - \frac{1}{x+6} = \frac{5}{28};\) 5) \(\frac{42}{x^2+5x} - \frac{3}{x^2-5x} = \frac{7}{x};\) 6) \(\frac{x}{x+8} + \frac{x+8}{x-8} = \frac{x^2+x+72}{x^2-64};\) 7) \(\frac{3x-5}{x^2-1} = \frac{3x+2}{x^2+x} - \frac{6x-5}{x^2-x};\)

Решение:

1) \(\frac{x^2-9x}{x+3} = \frac{36}{x+3};\) \(x^2 - 9x = 36\) \(x^2 - 9x - 36 = 0\) \(D = (-9)^2 - 4 \cdot 1 \cdot (-36) = 81 + 144 = 225\) \(x_1 = \frac{9 + \sqrt{225}}{2} = \frac{9 + 15}{2} = \frac{24}{2} = 12\) \(x_2 = \frac{9 - \sqrt{225}}{2} = \frac{9 - 15}{2} = \frac{-6}{2} = -3\) Но \(x
eq -3\), так как знаменатель не может быть равен нулю. 2) \(\frac{x^2+x}{x^2-25} = \frac{45-3x}{x^2-25};\) \(x^2 + x = 45 - 3x\) \(x^2 + 4x - 45 = 0\) \(D = 4^2 - 4 \cdot 1 \cdot (-45) = 16 + 180 = 196\) \(x_1 = \frac{-4 + \sqrt{196}}{2} = \frac{-4 + 14}{2} = \frac{10}{2} = 5\) \(x_2 = \frac{-4 - \sqrt{196}}{2} = \frac{-4 - 14}{2} = \frac{-18}{2} = -9\) Но \(x
eq 5\) и \(x
eq -5\), так как знаменатель не может быть равен нулю. 3) \(\frac{5x-8}{x-1} = \frac{14x+12}{3x+5};\) \((5x-8)(3x+5) = (14x+12)(x-1)\) \(15x^2 + 25x - 24x - 40 = 14x^2 - 14x + 12x - 12\) \(15x^2 + x - 40 = 14x^2 - 2x - 12\) \(x^2 + 3x - 28 = 0\) \(D = 3^2 - 4 \cdot 1 \cdot (-28) = 9 + 112 = 121\) \(x_1 = \frac{-3 + \sqrt{121}}{2} = \frac{-3 + 11}{2} = \frac{8}{2} = 4\) \(x_2 = \frac{-3 - \sqrt{121}}{2} = \frac{-3 - 11}{2} = \frac{-14}{2} = -7\) 4) \(\frac{1}{x-4} - \frac{1}{x+6} = \frac{5}{28};\) \(\frac{(x+6) - (x-4)}{(x-4)(x+6)} = \frac{5}{28};\) \(\frac{x+6-x+4}{x^2+6x-4x-24} = \frac{5}{28};\) \(\frac{10}{x^2+2x-24} = \frac{5}{28};\) \(5(x^2+2x-24) = 280\) \(5x^2+10x-120 = 280\) \(5x^2+10x-400 = 0\) \(x^2+2x-80 = 0\) \(D = 2^2 - 4 \cdot 1 \cdot (-80) = 4 + 320 = 324\) \(x_1 = \frac{-2 + \sqrt{324}}{2} = \frac{-2 + 18}{2} = \frac{16}{2} = 8\) \(x_2 = \frac{-2 - \sqrt{324}}{2} = \frac{-2 - 18}{2} = \frac{-20}{2} = -10\) 5) \(\frac{42}{x^2+5x} - \frac{3}{x^2-5x} = \frac{7}{x};\) \(\frac{42}{x(x+5)} - \frac{3}{x(x-5)} = \frac{7}{x};\) Умножаем обе части на \(x(x+5)(x-5)\): \(42(x-5) - 3(x+5) = 7(x+5)(x-5)\) \(42x - 210 - 3x - 15 = 7(x^2 - 25)\) \(39x - 225 = 7x^2 - 175\) \(7x^2 - 39x + 50 = 0\) \(D = (-39)^2 - 4 \cdot 7 \cdot 50 = 1521 - 1400 = 121\) \(x_1 = \frac{39 + \sqrt{121}}{14} = \frac{39 + 11}{14} = \frac{50}{14} = \frac{25}{7}\) \(x_2 = \frac{39 - \sqrt{121}}{14} = \frac{39 - 11}{14} = \frac{28}{14} = 2\) 6) \(\frac{x}{x+8} + \frac{x+8}{x-8} = \frac{x^2+x+72}{x^2-64};\) \(\frac{x(x-8) + (x+8)^2}{(x+8)(x-8)} = \frac{x^2+x+72}{x^2-64};\) \(x(x-8) + (x+8)^2 = x^2+x+72\) \(x^2 - 8x + x^2 + 16x + 64 = x^2 + x + 72\) \(2x^2 + 8x + 64 = x^2 + x + 72\) \(x^2 + 7x - 8 = 0\) \(D = 7^2 - 4 \cdot 1 \cdot (-8) = 49 + 32 = 81\) \(x_1 = \frac{-7 + \sqrt{81}}{2} = \frac{-7 + 9}{2} = \frac{2}{2} = 1\) \(x_2 = \frac{-7 - \sqrt{81}}{2} = \frac{-7 - 9}{2} = \frac{-16}{2} = -8\) Но \(x
eq -8\) и \(x
eq 8\), так как знаменатель не может быть равен нулю. 7) \(\frac{3x-5}{x^2-1} = \frac{3x+2}{x^2+x} - \frac{6x-5}{x^2-x};\) \(\frac{3x-5}{(x-1)(x+1)} = \frac{3x+2}{x(x+1)} - \frac{6x-5}{x(x-1)};\) \(\frac{3x-5}{(x-1)(x+1)} = \frac{(3x+2)(x-1) - (6x-5)(x+1)}{x(x+1)(x-1)};\) \((3x-5)x = (3x+2)(x-1) - (6x-5)(x+1)\) \(3x^2 - 5x = (3x^2 - 3x + 2x - 2) - (6x^2 + 6x - 5x - 5)\) \(3x^2 - 5x = 3x^2 - x - 2 - 6x^2 - x + 5\) \(3x^2 - 5x = -6x^2 + 3x + 3\) \(9x^2 - 8x - 3 = 0\) \(D = (-8)^2 - 4 \cdot 9 \cdot (-3) = 64 + 108 = 172\) \(x_1 = \frac{8 + \sqrt{172}}{18} = \frac{8 + 2\sqrt{43}}{18} = \frac{4 + \sqrt{43}}{9}\) \(x_2 = \frac{8 - \sqrt{172}}{18} = \frac{8 - 2\sqrt{43}}{18} = \frac{4 - \sqrt{43}}{9}\)

Ответ:

Ты молодец! У тебя всё получится!