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Автор - АЛЕНЬКЕЙЦЕТОЧЕК

тик трое поп жжжжжж​

Ответ

Автор - L0sK

№858.

а) frac{4x+1}{x-3} = frac{3x-8}{x+1}

(4x+1)(x+1) = (3x-8)(x-3)

4x² + 5x + 1 = 3x² - 17x + 24

x² + 22x - 23 = 0

По теореме Виета:

x₁ = 1; x₂ = -23

Ответ: -23; 1.

б) frac{x-2}{x+2} = frac{x+3}{x-4}

(x-2)(x-4) = (x+2)(x+3)

x² - 6x + 8 = x² + 5x + 6

11x = 2

x = frac{2}{11}

Ответ: frac{2}{11}.

№860.

a) frac{10}{(x-5)(x+1)} + frac{x}{x+1} = frac{3}{x-5}

frac{10 + x(x-5) - 3(x+1)}{(x-5)(x+1)} = 0

frac{10 + x^{2} - 5x - 3x -3}{(x-5)(x+1)} = 0

frac{x^{2} - 8x +7}{(x-5)(x+1)} = 0

x² - 8x +7 = 0          и            (x-5)(x+1) ≠ 0

По теореме Виета:             x ≠ 5; x ≠ -1

x₁ = 7; x₂ = 1

Ответ: 1; 7.

б) frac{36}{x(x-12)} - frac{3}{x-12} = 3

frac{36 - 3x - 3x(x-12)}{x(x-12)} = 0

frac{36 - 3x - 3x^{2} +36x}{x(x-12)} = 0

frac{3x^{2} - 33x - 36}{x(x-12)} = 0 | :3

frac{x^{2} - 11x - 12}{x(x-12)} = 0

x² - 11x - 12 = 0     и         x(x-12) ≠ 0

По теореме Виета:       x ≠ 0; x ≠ 12

x₁ = 12; x₂ = -1

Ответ: -1.

в) frac{2x-7}{x - 4} - frac{x+2}{x+1} = frac{x+6}{(x-4)(x+1)}

frac{x+6 - (2x-7)(x+1) + (x+2)(x-4)}{(x-4)(x+1)} = 0

frac{x + 6 - (2x^{2} -5x -7) + (x^{2} - 2x - 8)}{(x-4)(x+1)}\ = 0

frac{x + 6 -2x^{2} + 5x +7 + x^{2} - 2x - 8}{(x-4)(x+1)}\ = 0

frac{-x^{2} + 4x + 5 }{(x-4)(x+1)}\ = 0

frac{x^{2} - 4x - 5 }{(x-4)(x+1)}\ = 0

x² - 4x - 5 = 0        и         (x-4)(x+1) ≠ 0

По теореме Виета:        x ≠ 4; x ≠ -1

x₁ = 5; x₂ = -1

Ответ: 5.

г) frac{2x + 5}{x(x+1)} - frac{2}{x} - frac{3x}{x + 1} = 0

frac{2x + 5 - 2(x+1) - 3x*x}{x(x+1)} = 0

frac{2x + 5 - 2x -2 - 3x^{2} }{x(x+1)} = 0

frac{-3x^{2} + 3}{x(x+1)} = 0 | : (-3)

frac{x^{2} - 1}{x(x+1)} = 0

x² - 1 = 0           и     x(x+1) ≠ 0

x₁ = 1; x₂ = -1            x ≠ 0; x ≠ -1

Ответ: 1.

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